Deck 8: Essential Probability Rules

ملء الشاشة (f)
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سؤال
What is the probability that an event will occur?

A)The number of successful outcomes divided by the number of unsuccessful outcomes
B)The number of successful outcomes divided by the total number of outcomes
C)The number of unsuccessful outcomes divided by the number of successful outcomes
D)The number of unsuccessful outcomes divided by the total number of outcomes
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لقلب البطاقة.
سؤال
A population has an equal proportion of males and females. That is, when randomly selecting one individual, the probability that the individual is male (M) is ½ and the probability that the individual is female (F) is ½ Each of the outcomes M and F have a probability ½ of occurring. What does this mean?

A)In the next four randomly selected individuals, exactly two of the outcomes will be M.
B)In the next four randomly selected individuals, the outcomes will alternate between MFMF and FMFM.
C)In the next four randomly selected individuals, they cannot all be the same outcome.
D)In the next four randomly selected individuals, the outcomes could be any combination of M and F
سؤال
A population has an equal proportion of males and females. That is, when randomly selecting one individual, the probability that the individual is male (M) is ½ and the probability that the individual is female (F) is ½ There are only two outcomes when an individual is selected: {M, F}. What is this collection of all possible outcomes called?

A)A census
B)The population
C)The sample space
D)The distribution
سؤال
A population has an equal proportion of males and females. That is, when randomly selecting one individual, the probability that the individual is male (M) is ½ and the probability that the individual is female (F) is ½. In the first 50 randomly selected individuals, 20 were male. In the next 50 randomly selected individuals, which of the following must happen?

A)More than half of the individuals will be males to balance out the small number of males in the first 50 individuals. However, the order in which these males will be selected is unpredictable.
B)More than 20 of the individuals will be males because the proportion of males after 100 individuals must be closer to ½
C)The number of males will be very close to 30 in the next 50 individuals because the proportion must be close to ½
D)The number of males could be any value from 0 to 30 in the next 50 individuals because each individual is independent of the others.
سؤال
Studies indicate that approximately 7% of Americans are vegetarians. If an American is selected at random, what is the probability that he or she is NOT a vegetarian?
سؤال
Which of the following defines an event considered to have an unusual probability?

A)The probability exceeds 1.
B)The probability is a negative number.
C)The probability is very low.
D)The probability equals 1.
سؤال
A researcher is trying to determine the proportion of a certain species of fish in a local lake. After sampling 40 fish, she found 32 of them were the species of interest. What does she estimate as the probability that the next randomly selected fish will be of the species of interest?

A)0.32
B)0.50
C)0.80
D)1.25
سؤال
The Centers for Disease Control and Prevention receives information about the causes for HIV/AIDS infection in the United States. Here are the causes of infection and their respective probability among American women age 20 to 24 years:
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CauseInjection drug useHeterosexual sexPerinatal infectionOther causesProbability?0.6640.1250.095\begin{array}{c}\begin{array}{lll}\bf{Cause}\\\text{Injection drug use} \\\text{Heterosexual sex} \\\text{Perinatal infection} \\\text{Other causes}\end{array}\begin{array}{lll}\bf{Probability}\\? \\0.664 \\0.125 \\0.095\end{array}\end{array} This is a legitimate probability model. Therefore, what must be the probability that an American woman age 20 to 24 was infected via injection drug use?

A)0.116
B)0.186
C)0.206
D)0.250
سؤال
The Centers for Disease Control and Prevention receives information about the causes for HIV/AIDS infection in the United States. Here are the causes of infection and their respective probability among American women age 20 to 24 years:
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CauseInjection drug useHeterosexual sexPerinatal infectionOther causesProbability?0.6640.1250.095\begin{array}{c}\begin{array}{lll}\bf{Cause}\\\text{Injection drug use} \\\text{Heterosexual sex} \\\text{Perinatal infection} \\\text{Other causes}\end{array}\begin{array}{lll}\bf{Probability}\\? \\0.664 \\0.125 \\0.095\end{array}\end{array} What is the probability that an American woman age 20 to 24 contracted HIV/AIDS either via heterosexual sex or via perinatal infection?

A)0.083
B)0.211
C)0.500
D)0.789
سؤال
Which of the following terms describes an event that is uncertain but has a regular distribution over many outcomes?

A)Random
B)Predictable
C)Deterministic
D)Chaotic
سؤال
Event A has probability 0.4. Event B has probability 0.5. If A and B are disjoint, what is the probability that both events occur?

A)0.0
B)0.1
C)0.2
D)0.9
سؤال
Event A has probability 0.4. Event B has probability 0.5. If A and B are disjoint, what is the probability of A or B?

A)0.0
B)0.1
C)0.2
D)0.9
سؤال
Event A has probability 0.4. Event B has probability 0.5. Event C has probability 0.6. Which of the following statements about these events must be TRUE?

A)Events A, B, and C are disjoint.
B)One event must have some outcomes in common with at least one other event.
C)The probability that event C does not occur is 0.9.
D)The sample space for these events is larger than typical.
سؤال
Event A occurs with probability 0.2. Event B occurs with probability 0.3. Event C occurs with probability 0.4. If A, B, and C are disjoint, then which of the following probabilities is (are) correct?

A)P(A or B) = 0.5
B)P(A or C) = 0.6
C)P(A or B or C) = 0.9
D)All of these choices are correct.
سؤال
Based on a recent Gallup survey, we define the following probability model for the number X of daily cups of coffee consumed by Americans.
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X01234 or more  Probability 0.360.260.190.09?\begin{array} { | l | l | l | l | l | l | } \hline \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \text { or more } \\\hline \text { Probability } & 0.36 & 0.26 & 0.19 & 0.09 & ? \\\hline\end{array} What is the probability that an American consumes 4 or more cups of coffee daily?

A)0.10
B)1/5
C)1/4
D)Impossible to determine from the information given
سؤال
Based on a recent Gallup survey, we define the following probability model for the number X of daily cups of coffee consumed by Americans.


X01234 or more  Probability 0.360.260.190.09?\begin{array} { | l | l | l | l | l | l | } \hline \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \text { or more } \\\hline \text { Probability } & 0.36 & 0.26 & 0.19 & 0.09 & ? \\\hline\end{array} What is the probability that an American does NOT consume 2 cups of coffee daily?
سؤال
Based on a recent Gallup survey, we define the following probability model for the number X of daily cups of coffee consumed by Americans.


X01234 or more  Probability 0.360.260.190.09?\begin{array} { | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 \text { or more } \\\hline \text { Probability } & 0.36 & 0.26 & 0.19 & 0.09 & ? \\\hline\end{array} What is the probability that an American consumes at most 1 cup of coffee daily?
سؤال
What is the term for a variable whose value is a numerical outcome of a random phenomenon?

A)Random variable
B)Parameter
C)Biased
D)Random sample
سؤال
A physician observes the number of lesions on subjects who regularly used tanning salons. Let X be the number of lesions observed. The physician found that X had the following probability distribution.
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 Value of X01234 Probability 0.050.10.250.300.30\begin{array} { | l | l | l | l | l | l | } \hline \text { Value of } \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \\\hline \text { Probability } & 0.05 & 0.1 & 0.25 & 0.30 & 0.30 \\\hline\end{array} What is the probability that a randomly chosen subject has at least three lesions?

A)0.3
B)0.4
C)0.6
D)0.7
سؤال
A physician observes the number of lesions on subjects who regularly used tanning salons. Let X be the number of lesions observed. The physician found that X had the following probability distribution.
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 Value of X01234 Probability 0.050.10.250.300.30\begin{array} { | l | l | l | l | l | l | } \hline \text { Value of } \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \\\hline \text { Probability } & 0.05 & 0.1 & 0.25 & 0.30 & 0.30 \\\hline\end{array} What is the value of P(X > 3)?

A)0.3
B)0.4
C)0.6
D)0.7
سؤال
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
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 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} This is a legitimate probability model. Therefore, what is the value of P(X = 7)?

A)1/7
B)1/8
C)0.43
D)0.57
سؤال
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
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 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What does the event X < 5 mean?

A)The student watched TV no more than 5 days in the past week.
B)The student watched TV at least 5 days in the past week.
C)The probability that the student watched TV in the past week is smaller than 5.
D)The student watched TV fewer than 5 days in the past week.
سؤال
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
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 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the probability that X < 5?

A)0.08
B)1/8
C)0.30
D)0.38
سؤال
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
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 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the correct notation for "watched TV at least two days in the past week" in terms of X?

A)X > 1
B)X > 2
C)X = 2
D)X < 1
سؤال
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
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 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the probability that a randomly selected student watched TV at least two days in the past week?

A)2/7
B)5/8
C)0.87
D)0.93
سؤال
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
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 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the probability that a randomly selected student did not watch TV 7 days in the past week?

A)0.43
B)0.57
C)6/7
D)7/8
سؤال
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
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 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} For a randomly selected student, which of the following terms describes the events X = 3 and X = 4?

A)Disjoint
B)Not disjoint
C)Independent
D)Complementary
سؤال
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
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X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the probability of getting exactly 3 color-blind individuals in the sample?

A)0.000
B)0.019
C)0.111
D)0.125
سؤال
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
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X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the probability of getting at least 6 color-blind individuals in the sample?

A)0.000
B)0.333
C)0.375
D)1.000
سؤال
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
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X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the probability of getting 2 or 3 color-blind individuals in the sample?

A)0.039
B)0.128
C)0.466
D)0.870
سؤال
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
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X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} Which is the correct notation for representing the probability of getting at least 4 color-blind individuals in the sample?

A)P(X < 4)
B)P(X > 4)
C)P(X ?4)
D)P(X ? 4)
سؤال
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
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X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the mean of this distribution?

A)0.640
B)0.766
C)0.096
D)None of these choices is correct.
سؤال
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
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X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the standard deviation of this distribution?

A)0.64
B)0.766
C)0.096
D)None of these choices is correct.
سؤال
Based on data from the U.S. Department of Agriculture, we define the following probability model for the number X of different pesticides detected in fresh produce.


X012345 or more  Probability 0.430.170.140.080.060.12\begin{array} { | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 \text { or more } \\\hline \text { Probability } & 0.43 & 0.17 & 0.14 & 0.08 & 0.06 & 0.12 \\\hline\end{array} What is the numerical value for the probability P(X < 2)?
سؤال
Based on data from the U.S. Department of Agriculture, we define the following probability model for the number X of different pesticides detected in fresh produce.
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X012345 or more  Probability 0.430.170.140.080.060.12\begin{array} { | l | l | l | l | l | l | l | } \hline \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 & 5 \text { or more } \\\hline \text { Probability } & 0.43 & 0.17 & 0.14 & 0.08 & 0.06 & 0.12 \\\hline\end{array} In this context, the probability P(X < 2) is also equal to which of these probabilities?

A)P(X ? 2)
B)P(X ? 3)
C)P(X ? 1)
D)P(X = 1)
سؤال
If X is a continuous uniform random variable taking values between 0 and 100, what is the numerical value of the probability P(X = 10.333)?
سؤال
The expected value of a probability distribution is equal to

A)The standard deviation.
B)The mean.
C)The variance.
D)The value cannot be found because all the probabilities are different.
سؤال
The probability density of a random variable X is given in the figure.
From this density, what is the probability that X is between 0.5 and 1.5?
<strong>The probability density of a random variable X is given in the figure. From this density, what is the probability that X is between 0.5 and 1.5?  </strong> A)1/3 B)1/2 C)3/4 D)1 <div style=padding-top: 35px>

A)1/3
B)1/2
C)3/4
D)1
سؤال
The probability density of a random variable X is given in the figure.
What is the probability that X is at least 1.5?
<strong>The probability density of a random variable X is given in the figure. What is the probability that X is at least 1.5?  </strong> A)0 B)1/4 C)1/3 D)1/2 <div style=padding-top: 35px>

A)0
B)1/4
C)1/3
D)1/2
سؤال
The probability density of a random variable X is given in the figure.
What is the probability that X = 1.5?
<strong>The probability density of a random variable X is given in the figure. What is the probability that X = 1.5?  </strong> A)0 B)1/4 C)1/3 D)1/2 <div style=padding-top: 35px>

A)0
B)1/4
C)1/3
D)1/2
سؤال
The probability density of a random variable X is given in the figure.
What is the median of the distribution of X?
<strong>The probability density of a random variable X is given in the figure. What is the median of the distribution of X?  </strong> A)0 B)1 C)2 D)Impossible to calculate from the graph <div style=padding-top: 35px>

A)0
B)1
C)2
D)Impossible to calculate from the graph
سؤال
The probability density of a random variable X is given in the figure.
What is the value of P(0 < X ≤ 2)?

A)0.0
B)0.5
C)1.0
D)2.0
سؤال
Which of the following statements is NOT true about continuous probability distributions?

A)The probability of any event is the area under the density curve over the range of values that make up the event.
B)The total area under the density curve must be exactly 1.
C)If X is a continuous random variable taking values between 0 and 500, then P(X > 200) = P(X ≥ 200).
D)There are no disjoint events in continuous probability models.
سؤال
If X is a continuous uniform random variable taking values between 0 and 10, then the value of the probability P(X ≤ 4.3) is

A)Slightly smaller than the value of P(X < 4.3).
B)Exactly the same as the value of P(X < 4.3).
C)Slightly larger than the value of P(X < 4.3).
D)Possibly smaller or larger than the value of P(X < 4.3).
سؤال
A probability can be called a risk when it applies to an undesirable outcome.
سؤال
An odds is the ratio of the probability of an outcome over the probability of that outcome not
occurring.
سؤال
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
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 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} What is the probability that the person chosen has blood type A?

A)0.04
B)0.25
C)0.27
D)Impossible to determine from the information given
سؤال
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
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 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} What is the probability that the person chosen has a blood type other than O?

A)0.25
B)0.50
C)0.75
D)1.00
سؤال
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
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 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} Some people consider having type AB blood to be an undesirable outcome because it may be difficult to find matching donors. What is the risk of having type AB?

A)0.05
B)0.95
C)1.05
D)20.0
سؤال
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
?
?
 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} Considering having type AB blood to be an undesirable outcome, what are the odds of having type AB blood?

A)0.050
B)0.053
C)0.950
D)19.0
سؤال
According to the Centers for Disease Control and Prevention, as much as 20% of the U.S. population gets the flu each year. In light of this probability, what are the odds of catching the influenza virus?

A)0.20
B)0.25
C)0.80
D)4.00
سؤال
Many health professionals believe that attention disorders such as ADD and ADHD are being over-diagnosed. It is predicted that 60% of the diagnoses are incorrect. What are the odds of a diagnosis being incorrect?

A)0.40
B)0.60
C)0.67
D)1.50
سؤال
A recent nationwide study of myopia (nearsightedness) found that 38.1% of Americans age 18 to 24 suffer from myopia. What is the probability that a randomly chosen American between the ages of 18 and 24 does NOT have myopia?

A)-0.381
B)0.381
C)0.616
D)0.619
سؤال
A recent nationwide study of myopia (nearsightedness) found that 38.1% of Americans age 18 to 24 suffer from myopia. What are the odds of myopia in the 18- to 24-year-old American population?

A)0.381
B)0.616
C)1.616
D)2.625
سؤال
A recent nationwide study of myopia (nearsightedness) found that 38.1% of Americans age 18 to 24 suffer from myopia. What is the risk of myopia in the 18- to 24-year-old American population?

A)0.381
B)0.616
C)1.616
D)2.625
سؤال
The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among American women age 20 to 24 is 2791.5 per 100,000. What is the probability that a randomly chosen American woman between the ages of 20 and 24 does NOT have a Chlamydia infection?

A)0.0279
B)0.2792
C)0.7208
D)0.9721
سؤال
The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among American women age 20 to 24 is 2791.5 per 100,000. What is the approximate risk of a Chlamydia infection in a 20- to 24-year-old U.S. female?

A)0.0279
B)0.0287
C)0.2792
D)0.2872
سؤال
The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among American women age 20 to 24 is 2791.5 per 100,000. What are the approximate odds of a Chlamydia infection in the 20- to 24-year-old U.S. female population?

A)0.0279
B)0.0287
C)0.2792
D)0.2872
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Deck 8: Essential Probability Rules
1
What is the probability that an event will occur?

A)The number of successful outcomes divided by the number of unsuccessful outcomes
B)The number of successful outcomes divided by the total number of outcomes
C)The number of unsuccessful outcomes divided by the number of successful outcomes
D)The number of unsuccessful outcomes divided by the total number of outcomes
B
2
A population has an equal proportion of males and females. That is, when randomly selecting one individual, the probability that the individual is male (M) is ½ and the probability that the individual is female (F) is ½ Each of the outcomes M and F have a probability ½ of occurring. What does this mean?

A)In the next four randomly selected individuals, exactly two of the outcomes will be M.
B)In the next four randomly selected individuals, the outcomes will alternate between MFMF and FMFM.
C)In the next four randomly selected individuals, they cannot all be the same outcome.
D)In the next four randomly selected individuals, the outcomes could be any combination of M and F
D
3
A population has an equal proportion of males and females. That is, when randomly selecting one individual, the probability that the individual is male (M) is ½ and the probability that the individual is female (F) is ½ There are only two outcomes when an individual is selected: {M, F}. What is this collection of all possible outcomes called?

A)A census
B)The population
C)The sample space
D)The distribution
C
4
A population has an equal proportion of males and females. That is, when randomly selecting one individual, the probability that the individual is male (M) is ½ and the probability that the individual is female (F) is ½. In the first 50 randomly selected individuals, 20 were male. In the next 50 randomly selected individuals, which of the following must happen?

A)More than half of the individuals will be males to balance out the small number of males in the first 50 individuals. However, the order in which these males will be selected is unpredictable.
B)More than 20 of the individuals will be males because the proportion of males after 100 individuals must be closer to ½
C)The number of males will be very close to 30 in the next 50 individuals because the proportion must be close to ½
D)The number of males could be any value from 0 to 30 in the next 50 individuals because each individual is independent of the others.
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5
Studies indicate that approximately 7% of Americans are vegetarians. If an American is selected at random, what is the probability that he or she is NOT a vegetarian?
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6
Which of the following defines an event considered to have an unusual probability?

A)The probability exceeds 1.
B)The probability is a negative number.
C)The probability is very low.
D)The probability equals 1.
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7
A researcher is trying to determine the proportion of a certain species of fish in a local lake. After sampling 40 fish, she found 32 of them were the species of interest. What does she estimate as the probability that the next randomly selected fish will be of the species of interest?

A)0.32
B)0.50
C)0.80
D)1.25
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8
The Centers for Disease Control and Prevention receives information about the causes for HIV/AIDS infection in the United States. Here are the causes of infection and their respective probability among American women age 20 to 24 years:
?
?
CauseInjection drug useHeterosexual sexPerinatal infectionOther causesProbability?0.6640.1250.095\begin{array}{c}\begin{array}{lll}\bf{Cause}\\\text{Injection drug use} \\\text{Heterosexual sex} \\\text{Perinatal infection} \\\text{Other causes}\end{array}\begin{array}{lll}\bf{Probability}\\? \\0.664 \\0.125 \\0.095\end{array}\end{array} This is a legitimate probability model. Therefore, what must be the probability that an American woman age 20 to 24 was infected via injection drug use?

A)0.116
B)0.186
C)0.206
D)0.250
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9
The Centers for Disease Control and Prevention receives information about the causes for HIV/AIDS infection in the United States. Here are the causes of infection and their respective probability among American women age 20 to 24 years:
?
?
CauseInjection drug useHeterosexual sexPerinatal infectionOther causesProbability?0.6640.1250.095\begin{array}{c}\begin{array}{lll}\bf{Cause}\\\text{Injection drug use} \\\text{Heterosexual sex} \\\text{Perinatal infection} \\\text{Other causes}\end{array}\begin{array}{lll}\bf{Probability}\\? \\0.664 \\0.125 \\0.095\end{array}\end{array} What is the probability that an American woman age 20 to 24 contracted HIV/AIDS either via heterosexual sex or via perinatal infection?

A)0.083
B)0.211
C)0.500
D)0.789
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10
Which of the following terms describes an event that is uncertain but has a regular distribution over many outcomes?

A)Random
B)Predictable
C)Deterministic
D)Chaotic
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11
Event A has probability 0.4. Event B has probability 0.5. If A and B are disjoint, what is the probability that both events occur?

A)0.0
B)0.1
C)0.2
D)0.9
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12
Event A has probability 0.4. Event B has probability 0.5. If A and B are disjoint, what is the probability of A or B?

A)0.0
B)0.1
C)0.2
D)0.9
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13
Event A has probability 0.4. Event B has probability 0.5. Event C has probability 0.6. Which of the following statements about these events must be TRUE?

A)Events A, B, and C are disjoint.
B)One event must have some outcomes in common with at least one other event.
C)The probability that event C does not occur is 0.9.
D)The sample space for these events is larger than typical.
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14
Event A occurs with probability 0.2. Event B occurs with probability 0.3. Event C occurs with probability 0.4. If A, B, and C are disjoint, then which of the following probabilities is (are) correct?

A)P(A or B) = 0.5
B)P(A or C) = 0.6
C)P(A or B or C) = 0.9
D)All of these choices are correct.
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15
Based on a recent Gallup survey, we define the following probability model for the number X of daily cups of coffee consumed by Americans.
?
?
X01234 or more  Probability 0.360.260.190.09?\begin{array} { | l | l | l | l | l | l | } \hline \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \text { or more } \\\hline \text { Probability } & 0.36 & 0.26 & 0.19 & 0.09 & ? \\\hline\end{array} What is the probability that an American consumes 4 or more cups of coffee daily?

A)0.10
B)1/5
C)1/4
D)Impossible to determine from the information given
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16
Based on a recent Gallup survey, we define the following probability model for the number X of daily cups of coffee consumed by Americans.


X01234 or more  Probability 0.360.260.190.09?\begin{array} { | l | l | l | l | l | l | } \hline \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \text { or more } \\\hline \text { Probability } & 0.36 & 0.26 & 0.19 & 0.09 & ? \\\hline\end{array} What is the probability that an American does NOT consume 2 cups of coffee daily?
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17
Based on a recent Gallup survey, we define the following probability model for the number X of daily cups of coffee consumed by Americans.


X01234 or more  Probability 0.360.260.190.09?\begin{array} { | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 \text { or more } \\\hline \text { Probability } & 0.36 & 0.26 & 0.19 & 0.09 & ? \\\hline\end{array} What is the probability that an American consumes at most 1 cup of coffee daily?
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18
What is the term for a variable whose value is a numerical outcome of a random phenomenon?

A)Random variable
B)Parameter
C)Biased
D)Random sample
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19
A physician observes the number of lesions on subjects who regularly used tanning salons. Let X be the number of lesions observed. The physician found that X had the following probability distribution.
?
?
 Value of X01234 Probability 0.050.10.250.300.30\begin{array} { | l | l | l | l | l | l | } \hline \text { Value of } \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \\\hline \text { Probability } & 0.05 & 0.1 & 0.25 & 0.30 & 0.30 \\\hline\end{array} What is the probability that a randomly chosen subject has at least three lesions?

A)0.3
B)0.4
C)0.6
D)0.7
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20
A physician observes the number of lesions on subjects who regularly used tanning salons. Let X be the number of lesions observed. The physician found that X had the following probability distribution.
?
?
 Value of X01234 Probability 0.050.10.250.300.30\begin{array} { | l | l | l | l | l | l | } \hline \text { Value of } \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 \\\hline \text { Probability } & 0.05 & 0.1 & 0.25 & 0.30 & 0.30 \\\hline\end{array} What is the value of P(X > 3)?

A)0.3
B)0.4
C)0.6
D)0.7
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21
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
?
?
 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} This is a legitimate probability model. Therefore, what is the value of P(X = 7)?

A)1/7
B)1/8
C)0.43
D)0.57
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22
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
?
?
 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What does the event X < 5 mean?

A)The student watched TV no more than 5 days in the past week.
B)The student watched TV at least 5 days in the past week.
C)The probability that the student watched TV in the past week is smaller than 5.
D)The student watched TV fewer than 5 days in the past week.
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23
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
?
?
 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the probability that X < 5?

A)0.08
B)1/8
C)0.30
D)0.38
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24
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
?
?
 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the correct notation for "watched TV at least two days in the past week" in terms of X?

A)X > 1
B)X > 2
C)X = 2
D)X < 1
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25
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
?
?
 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the probability that a randomly selected student watched TV at least two days in the past week?

A)2/7
B)5/8
C)0.87
D)0.93
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26
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
?
?
 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} What is the probability that a randomly selected student did not watch TV 7 days in the past week?

A)0.43
B)0.57
C)6/7
D)7/8
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27
Some argue that millions of Americans are so hooked on television that their viewing habits fit the criteria for substance abuse, as defined in the official psychiatric manual. Choose a young person (age 19 to 25) at random and ask, "In the past seven days, how many days did you watch television?" Call the response X for short. Here is a probability model for the response:
?
?
 Days X01234567 Probability 0.040.030.060.080.090.080.05?\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Days } X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Probability } & 0.04 & 0.03 & 0.06 & 0.08 & 0.09 & 0.08 & 0.05 & ? \\\hline\end{array} For a randomly selected student, which of the following terms describes the events X = 3 and X = 4?

A)Disjoint
B)Not disjoint
C)Independent
D)Complementary
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28
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
?
?
X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the probability of getting exactly 3 color-blind individuals in the sample?

A)0.000
B)0.019
C)0.111
D)0.125
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29
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
?
?
X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the probability of getting at least 6 color-blind individuals in the sample?

A)0.000
B)0.333
C)0.375
D)1.000
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30
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
?
?
X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the probability of getting 2 or 3 color-blind individuals in the sample?

A)0.039
B)0.128
C)0.466
D)0.870
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31
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
?
?
X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} Which is the correct notation for representing the probability of getting at least 4 color-blind individuals in the sample?

A)P(X < 4)
B)P(X > 4)
C)P(X ?4)
D)P(X ? 4)
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32
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
?
?
X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the mean of this distribution?

A)0.640
B)0.766
C)0.096
D)None of these choices is correct.
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33
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
?
?
X012345678P(X)0.5130.3570.1090.0190.0020.0000.0000.0000.000\begin{array} { | l | l | l | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \\\hline\end{array} What is the standard deviation of this distribution?

A)0.64
B)0.766
C)0.096
D)None of these choices is correct.
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34
Based on data from the U.S. Department of Agriculture, we define the following probability model for the number X of different pesticides detected in fresh produce.


X012345 or more  Probability 0.430.170.140.080.060.12\begin{array} { | l | l | l | l | l | l | l | } \hline X & 0 & 1 & 2 & 3 & 4 & 5 \text { or more } \\\hline \text { Probability } & 0.43 & 0.17 & 0.14 & 0.08 & 0.06 & 0.12 \\\hline\end{array} What is the numerical value for the probability P(X < 2)?
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35
Based on data from the U.S. Department of Agriculture, we define the following probability model for the number X of different pesticides detected in fresh produce.
?
X012345 or more  Probability 0.430.170.140.080.060.12\begin{array} { | l | l | l | l | l | l | l | } \hline \boldsymbol { X } & 0 & 1 & 2 & 3 & 4 & 5 \text { or more } \\\hline \text { Probability } & 0.43 & 0.17 & 0.14 & 0.08 & 0.06 & 0.12 \\\hline\end{array} In this context, the probability P(X < 2) is also equal to which of these probabilities?

A)P(X ? 2)
B)P(X ? 3)
C)P(X ? 1)
D)P(X = 1)
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36
If X is a continuous uniform random variable taking values between 0 and 100, what is the numerical value of the probability P(X = 10.333)?
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37
The expected value of a probability distribution is equal to

A)The standard deviation.
B)The mean.
C)The variance.
D)The value cannot be found because all the probabilities are different.
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38
The probability density of a random variable X is given in the figure.
From this density, what is the probability that X is between 0.5 and 1.5?
<strong>The probability density of a random variable X is given in the figure. From this density, what is the probability that X is between 0.5 and 1.5?  </strong> A)1/3 B)1/2 C)3/4 D)1

A)1/3
B)1/2
C)3/4
D)1
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39
The probability density of a random variable X is given in the figure.
What is the probability that X is at least 1.5?
<strong>The probability density of a random variable X is given in the figure. What is the probability that X is at least 1.5?  </strong> A)0 B)1/4 C)1/3 D)1/2

A)0
B)1/4
C)1/3
D)1/2
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40
The probability density of a random variable X is given in the figure.
What is the probability that X = 1.5?
<strong>The probability density of a random variable X is given in the figure. What is the probability that X = 1.5?  </strong> A)0 B)1/4 C)1/3 D)1/2

A)0
B)1/4
C)1/3
D)1/2
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41
The probability density of a random variable X is given in the figure.
What is the median of the distribution of X?
<strong>The probability density of a random variable X is given in the figure. What is the median of the distribution of X?  </strong> A)0 B)1 C)2 D)Impossible to calculate from the graph

A)0
B)1
C)2
D)Impossible to calculate from the graph
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42
The probability density of a random variable X is given in the figure.
What is the value of P(0 < X ≤ 2)?

A)0.0
B)0.5
C)1.0
D)2.0
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43
Which of the following statements is NOT true about continuous probability distributions?

A)The probability of any event is the area under the density curve over the range of values that make up the event.
B)The total area under the density curve must be exactly 1.
C)If X is a continuous random variable taking values between 0 and 500, then P(X > 200) = P(X ≥ 200).
D)There are no disjoint events in continuous probability models.
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44
If X is a continuous uniform random variable taking values between 0 and 10, then the value of the probability P(X ≤ 4.3) is

A)Slightly smaller than the value of P(X < 4.3).
B)Exactly the same as the value of P(X < 4.3).
C)Slightly larger than the value of P(X < 4.3).
D)Possibly smaller or larger than the value of P(X < 4.3).
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45
A probability can be called a risk when it applies to an undesirable outcome.
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46
An odds is the ratio of the probability of an outcome over the probability of that outcome not
occurring.
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47
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
?
?
 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} What is the probability that the person chosen has blood type A?

A)0.04
B)0.25
C)0.27
D)Impossible to determine from the information given
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48
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
?
 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} What is the probability that the person chosen has a blood type other than O?

A)0.25
B)0.50
C)0.75
D)1.00
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49
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
?
?
 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} Some people consider having type AB blood to be an undesirable outcome because it may be difficult to find matching donors. What is the risk of having type AB?

A)0.05
B)0.95
C)1.05
D)20.0
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50
All human blood can be typed as either O, A, B, or AB. The distribution of the types varies a bit with race. Choose an African American person at random. Here are the approximate probabilities that the person you choose will have blood type O, B, or AB:
?
?
 Blood type  O  A  B  AB  Probability 0.50?0.200.05\begin{array} { | l | l | l | l | l | } \hline \text { Blood type } & \text { O } & \text { A } & \text { B } & \text { AB } \\\hline \text { Probability } & 0.50 & ? & 0.20 & 0.05 \\\hline\end{array} Considering having type AB blood to be an undesirable outcome, what are the odds of having type AB blood?

A)0.050
B)0.053
C)0.950
D)19.0
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51
According to the Centers for Disease Control and Prevention, as much as 20% of the U.S. population gets the flu each year. In light of this probability, what are the odds of catching the influenza virus?

A)0.20
B)0.25
C)0.80
D)4.00
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52
Many health professionals believe that attention disorders such as ADD and ADHD are being over-diagnosed. It is predicted that 60% of the diagnoses are incorrect. What are the odds of a diagnosis being incorrect?

A)0.40
B)0.60
C)0.67
D)1.50
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53
A recent nationwide study of myopia (nearsightedness) found that 38.1% of Americans age 18 to 24 suffer from myopia. What is the probability that a randomly chosen American between the ages of 18 and 24 does NOT have myopia?

A)-0.381
B)0.381
C)0.616
D)0.619
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54
A recent nationwide study of myopia (nearsightedness) found that 38.1% of Americans age 18 to 24 suffer from myopia. What are the odds of myopia in the 18- to 24-year-old American population?

A)0.381
B)0.616
C)1.616
D)2.625
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55
A recent nationwide study of myopia (nearsightedness) found that 38.1% of Americans age 18 to 24 suffer from myopia. What is the risk of myopia in the 18- to 24-year-old American population?

A)0.381
B)0.616
C)1.616
D)2.625
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56
The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among American women age 20 to 24 is 2791.5 per 100,000. What is the probability that a randomly chosen American woman between the ages of 20 and 24 does NOT have a Chlamydia infection?

A)0.0279
B)0.2792
C)0.7208
D)0.9721
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57
The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among American women age 20 to 24 is 2791.5 per 100,000. What is the approximate risk of a Chlamydia infection in a 20- to 24-year-old U.S. female?

A)0.0279
B)0.0287
C)0.2792
D)0.2872
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58
The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among American women age 20 to 24 is 2791.5 per 100,000. What are the approximate odds of a Chlamydia infection in the 20- to 24-year-old U.S. female population?

A)0.0279
B)0.0287
C)0.2792
D)0.2872
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