Question 1

Solve the problem.
-If the odds against a candidate winning an election are 7 to 6, then what is the probability that that candidate will win the election? 
A) $\frac { 7 } { 6 }$
B) $\frac { 7 } { 13 }$
C) $\frac { 6 } { 13 }$
D) $\frac { 6 } { 7 }$

Accepted Answer

The odds against winning are 7 to 6, meaning for every 7 chances of losing, there are 6 chances of winning. The total number of outcomes is 7 + 6 = 13. Therefore, the probability of winning is the number of winning outcomes over the total number of outcomes, which is $\frac{6}{13}$.

Question 2

Solve the problem.
-The kings are separated from a deck of standard playing cards and shuffled. One is randomly selected. What are the odds of drawing a black card?
A)4 : 1
B)1 : 4
C)1 : 2
D)1 : 1

Accepted Answer

In a standard deck, there are 2 black kings (spades and clubs) and 2 red kings (hearts and diamonds). The odds of drawing a black king are 2 black kings to 2 red kings, which simplifies to 1:1.

Question 3

Solve the problem.
-A basketball player with a free throw shooting percentage of 80% gets fouled and goes to the line to shoot two free throws. What are the odds of her making her first shot?
A)5 : 4
B)4 : 5
C)1 : 4
D)4 : 1

Accepted Answer

The answer of Solve the problem.
-A basketball player with a...

Question 4

Solve the problem.
-Of the coffee makers sold in an appliance store, 5% have either a faulty switch or a defective cord, 1.3% have a faulty switch, and 0.3% have both defects. What is the probability that a coffee maker will have a defective cord given that it has either a faulty switch or defective cord? Round to the nearest thousandth.
A)0.003
B)0.013
C)0.260
D)0.800

Accepted Answer

To find the probability that a coffee maker has a defective cord given it has either a faulty switch or a defective cord, we use the formula for conditional probability. The probability of having a defective cord (C) given either a faulty switch (S) or a defective cord is P(C|S or C) = P(C and (S or C)) / P(S or C). We know that P(S or C) = 5%, P(S) = 1.3%, P(C and S) = 0.3%. To find P(C), we use the principle of inclusion-exclusion: P(S or C) = P(S) + P(C) - P(S and C), so 5% = 1.3% + P(C) - 0.3%, which gives P(C) = 4%. The probability of having a defective cord given either fault is P(C|S or C) = P(C) / P(S or C) = 4% / 5% = 0.8 or 80%, which rounds to 0.800.

Question 5

Solve the problem.
-$\frac { 1 } { 25 },$ the odds against a direct hit?
A)25 : 1
B)1: 25
C)24 : 1
D)23 : 1

Accepted Answer

Odds against an event = (Probability of failure) : (Probability of success). Given the probability of a direct hit is $\frac{1}{25}$, the probability of failure is $1 - \frac{1}{25} = \frac{24}{25}$. Therefore, the odds against a direct hit are 24:1.

Question 6

Solve the problem.
-If you are dealt two cards successively (without replacement)from a standard 52-card deck, find the probability of getting a king on the first card and a queen on the second. 
A) $\frac { 4 } { 663 }$
B) $\frac { 13 } { 102 }$
C) $\frac { 1 } { 663 }$
D) $\frac { 2 } { 13 }$

Accepted Answer

The probability of drawing a king first is $ \frac{4}{52} $ since there are 4 kings in a 52-card deck. After drawing a king, there are 51 cards left, and the probability of then drawing a queen is $ \frac{4}{51} $ since there are 4 queens left. Multiplying these probabilities gives $ \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663} $.

Question 7

Solve the problem.
-A survey of senior citizens at a doctor's office shows that 53% of the seniors take blood pressure lowering medication and 47% take cholesterol lowering medication. 6% take both medications. What is the probability that a senior citizen takes only one of these medications given that he or she takes at least one of the medications?
A)0.53
B)0.47
C)-0.060
D)0.940

Accepted Answer

The probability that a senior citizen takes only one of these medications given that he or she takes at least one of the medications can be found by subtracting the probability of taking both medications from the total probability of taking either medication. The total probability of taking either medication is the sum of the probabilities of taking each medication minus the probability of taking both (to avoid double counting), which is 0.53 + 0.47 - 0.06 = 0.94 or 94%.

Question 8

Solve the problem.
-Seven slips of paper marked with the numbers 1, 2, 3, 4, 5, 6, and 7 are placed in a box and mixed well. Two are drawn. What are the odds that the sum of the numbers on the two selected slips is not 5?
A)9 : 1
B)21 : 2
C)19 :1
D)19 :2

Accepted Answer

The total number of ways to draw 2 slips out of 7 is C(7,2) = 21. The combinations that sum to 5 are (1,4) and (2,3), which are 2 in total. Therefore, the number of draws that do not sum to 5 is 21 - 2 = 19. The odds in favor of not summing to 5 are 19 to 2.

Question 9

Solve the problem.
-A survey revealed that 47% of people are entertained by reading books, 27% by watching TV, and 14% are entertained by both books and TV. What is the probability that a person will be entertained by books given that the person is entertained by TV? Round to the nearest thousandth.
A)0.27
B)0.519
C)0.298
D)0.47

Accepted Answer

The probability that a person will be entertained by books given that the person is entertained by TV is found using the formula for conditional probability: P(A|B) = P(A and B) / P(B). Here, A is the event of being entertained by books, and B is the event of being entertained by TV. P(A and B) is 14%, and P(B) is 27%. So, P(A|B) = 0.14 / 0.27 = 0.5185, which rounds to 0.519.

Question 10

Solve the problem.
-Seven slips of paper marked with the numbers 1, 2, 3, 4, 5, 6, and 7 are placed in a box and mixed well. Two are drawn. What are the odds that the sum of the numbers on the two selected slips is 8?
A)2 to 21
B)1 : 7
C)1 to 9
D)1 to 6

Accepted Answer

There are 21 possible pairs of numbers (since 7 choose 2 is 21). The pairs that sum to 8 are (1,7), (2,6), and (3,5), which are 3 pairs. The odds are calculated as the number of successful outcomes to unsuccessful ones, which is 3 to 18, simplifying to 1 to 6.

Question 11

Solve the problem.
-If you are dealt two cards successively (without replacement)from a standard 52-card deck, find the probability of getting a black on the first card and a heart on the second. 
A) $\frac { 13 } { 102 }$
B) $\frac { 13 } { 51 }$
C) $\frac { 7 } { 51 }$
D) $\frac { 1 } { 2,652 }$

Accepted Answer

The probability of drawing a black card first is $\frac{26}{52}$ since half of the 52 cards are black. After drawing a black card, there are 51 cards left, and 13 of them are hearts. Thus, the probability of then drawing a heart is $\frac{13}{51}$. The combined probability is the product of these two probabilities: $\frac{26}{52} \times \frac{13}{51} = \frac{1}{2} \times \frac{13}{51} = \frac{13}{102}$.

Question 12

Solve the problem.
-A football player with a field goal kicking percentage of 60% for kicks of 40 yards or less attempts a final-minute field goal of 32 yards. What are the odds of a successful kick?
A)5 : 3
B)2 : 3
C)3 : 5
D)3 : 2

Accepted Answer

The probability of a successful kick is 60%, which means the odds in favor of a successful kick are 60% to 40%. Converting this to odds, we have 60:40, which simplifies to 3:2.

Question 13

Solve the problem.
-A baseball player with a batting average of 0.400 comes to bat. What are the odds of his getting a hit?
A)2 : 5
B)5 : 2
C)2 : 3
D)3 : 2

Accepted Answer

The odds of getting a hit are calculated as the ratio of the probability of getting a hit to the probability of not getting a hit. With a batting average of 0.400, the probability of getting a hit is 0.4, and the probability of not getting a hit is 0.6. The odds are 0.4 : 0.6, which simplifies to 2 : 3.

Question 14

Solve the problem.
-If the odds that it will rain are 1 to 8, what is the probability of rain? 
A) $\frac { 1 } { 8 }$
B) $\frac { 1 } { 9 }$
C) $\frac { 8 } { 9 }$
D) $8 : 9$

Accepted Answer

The probability of an event is given by the ratio of the favorable outcomes to the total number of outcomes. Here, if the odds of rain are 1 to 8, it means there is 1 favorable outcome (rain) for every 8 unfavorable outcomes (no rain), making a total of 9 outcomes. Therefore, the probability of rain is $\frac{1}{9}$.

Question 15

Solve the problem.
-$\frac { 1 } { 25 } ,$ what are the odds against an earthquake?
A)23 : 1
B)25 : 1
C)24 : 1
D)1: 25

Accepted Answer

Odds against an event = (Probability of not happening) : (Probability of happening). Given the probability of an earthquake is $\frac{1}{25}$, the probability of not happening is $1 - \frac{1}{25} = \frac{24}{25}$. Thus, the odds against are 24:1.

Quiz 9: Sequences and Series; Counting and Probability

Solve the problem. -If the odds against a candidate winning an election are 7 to 6, then what is the probability that that candidate will win the election?

Solve the problem. -The kings are separated from a deck of standard playing cards and shuffled. One is randomly selected. What are the odds of drawing a black card?

Solve the problem. -A basketball player with a free throw shooting percentage of 80% gets fouled and goes to the line to shoot two free throws. What are the odds of her making her first shot?

Solve the problem. - $\frac { 1 } { 25 },$ the odds against a direct hit?

Solve the problem. -If you are dealt two cards successively (without replacement) from a standard 52-card deck, find the probability of getting a king on the first card and a queen on the second.

Solve the problem. -Seven slips of paper marked with the numbers 1, 2, 3, 4, 5, 6, and 7 are placed in a box and mixed well. Two are drawn. What are the odds that the sum of the numbers on the two selected slips is not 5?

Solve the problem. -A survey revealed that 47% of people are entertained by reading books, 27% by watching TV, and 14% are entertained by both books and TV. What is the probability that a person will be entertained by books given that the person is entertained by TV? Round to the nearest thousandth.

Solve the problem. -Seven slips of paper marked with the numbers 1, 2, 3, 4, 5, 6, and 7 are placed in a box and mixed well. Two are drawn. What are the odds that the sum of the numbers on the two selected slips is 8?

Solve the problem. -If you are dealt two cards successively (without replacement) from a standard 52-card deck, find the probability of getting a black on the first card and a heart on the second.

Solve the problem. -A football player with a field goal kicking percentage of 60% for kicks of 40 yards or less attempts a final-minute field goal of 32 yards. What are the odds of a successful kick?

Solve the problem. -A baseball player with a batting average of 0.400 comes to bat. What are the odds of his getting a hit?

Solve the problem. -If the odds that it will rain are 1 to 8, what is the probability of rain?

Solve the problem. - $\frac { 1 } { 25 } ,$ what are the odds against an earthquake?