Question 1

Compute Probabilities Involving Permutations and Combinations
-If you are dealt 6 cards from a shuffled deck of 52 cards, find the probability of getting 3 jacks and 3 aces.
A) $\frac { 2 } { 2,544,815 }$
B) $\frac { 2 } { 13 }$
C) $\frac { 1 } { 1,017,926 }$
D) $\frac { 3 } { 26 }$

Accepted Answer

The answer of Compute Probabilities Involving Permutations and Combinations
-If you...

Question 2

Determine the Appropriate Probability Rule to Use
-Find $P ( A$ or $B )$ given that $P ( A ) = 0.7 , P ( B ) = 0.2$, and $A$ and $B$ are mutually exclusive.
A) $0.9$
B) 0
C) $0.14$
D) $0.5$

Accepted Answer

Since $A$ and $B$ are mutually exclusive, $P(A \text{ or } B) = P(A) + P(B) = 0.7 + 0.2 = 0.9$.

Question 3

Solve Counting Problems Involving Permutations with Nondistinct Items
-How many distinct arrangements can be formed from all the letters of statistics
A) 50,400
B) 201,600
C) 259,200
D) 72
E) 3,628,800

Accepted Answer

The answer of Solve Counting Problems Involving Permutations with Nondistinct...

Question 4

Compute Probabilities Involving Permutations and Combinations
-A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 8 members and any member can be elected to each position No person can hold more than one office.
A) 336
B) 168
C) 112
D) 1680

Accepted Answer

Since no person can hold more than one office and there are 3 positions to fill from 8 members, we use permutations because the order matters (different positions). The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of items, and r is the number of items to choose. Here, n = 8 (members) and r = 3 (positions), so P(8, 3) = 8! / (8-3)! = 8! / 5! = (8×7×6×5!) / 5! = 8×7×6 = 336.

Question 5

Solve Counting Problems Involving Permutations with Nondistinct Items
-How many distinct arrangements of the letters in the word Mississippi are possible
A) 34,650
B) 39,916,800
C) 1152
D) 7920

Accepted Answer

The answer of Solve Counting Problems Involving Permutations with Nondistinct...

Question 6

Solve Counting Problems Involving Permutations with Nondistinct Items
-A man has 12 coins that consist of 3 pennies, 4 nickels, and 5 quarters. How many distinct arrangements of the coins can he make if he lays them in a row one at a time
A) 27,720
B) 479,001,600
C) 9240
D) 83,160

Accepted Answer

The answer of Solve Counting Problems Involving Permutations with Nondistinct...

Question 7

Compute Probabilities Involving Permutations and Combinations
-If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that all 5 cards are picture cards.
A) $\frac { 33 } { 108,290 }$
B) $\frac { 3 } { 13 }$
C) $\frac { 1 } { 2,598,960 }$
D) $\frac { 1 } { 216,580 }$

Accepted Answer

The total number of ways to choose 5 cards from a 52-card deck is $\binom{52}{5}$. There are 12 picture cards in a deck (Jacks, Queens, Kings of each suit), so the number of ways to choose 5 picture cards is $\binom{12}{5}$. The probability is $\frac{\binom{12}{5}}{\binom{52}{5}} = \frac{792}{2,598,960} = \frac{33}{108,290}$.

Question 8

Compute Probabilities Involving Permutations and Combinations
-Six students, A, B, C, D, E, F, are to give speeches to the class. The order of speaking is determined by random selection. Find the probability that (a) E will speak first (b) that $C$ will speak fifth and $B$ will speak last (c) that the students will speak in the following order: DECABF (d) that A or B will speak first.
A) $\frac { 1 } { 6 } ; \frac { 1 } { 24 } ; \frac { 1 } { 720 } ; \frac { 1 } { 3 }$
B) $\frac { 1 } { 6 } ; \frac { 1 } { 12 } ; \frac { 1 } { 720 } ; \frac { 1 } { 3 }$
C) $\frac { 1 } { 6 } ; \frac { 1 } { 36 } ; \frac { 1 } { 720 } ; \frac { 1 } { 12 }$
D) $\frac { 1 } { 6 } ; \frac { 1 } { 36 } ; \frac { 1 } { 360 } ; \frac { 1 } { 3 }$

Accepted Answer

(a) The probability E will speak first is $1/6$ because there are 6 students and each has an equal chance of being selected first. (b) The probability C will speak fifth and B will speak last is $1/(6*5) = 1/30$ because once C is chosen for the fifth position, there are 5 choices left for the last position, but since B must be last, it's actually just 1 way out of the 30 possible arrangements of the remaining students after choosing the first 4. However, this calculation is incorrect as the correct approach should consider all 6 positions, leading to a $1/6$ chance for C to be fifth and then a $1/5$ chance for B to be last among the remaining, giving $1/(6*5) = 1/30$, but since there's no matching option, we proceed with the provided options. (c) The probability that the students will speak in the order DECABF is $1/720$ because there is only 1 way this can happen out of the 720 possible arrangements (6!). (d) The probability that A or B will speak first is $2/6 = 1/3$ because there are 2 favorable outcomes out of 6 possible students who could be chosen first. The correct explanation for (b) should acknowledge that with 6 positions to fill, choosing C for the fifth and B for the last is a specific arrangement among all possible arrangements, leading to $1/(6*5) = 1/30$, but the correct calculation is $1/6$ for C's position times $1/5$ for B's position among the remaining, which simplifies to $1/30$, aligning with the principle of calculating probabilities for specific positions in a sequence. However, given the options and the need to correct the approach, the focus is on matching the provided answers and acknowledging the mistake in the initial explanation.

Question 9

Compute Probabilities Involving Permutations and Combinations
-In how many ways can a committee of three men and four women be formed from a group of 11 men and 11 women
A) 54450
B) 7,840,800
C) 554400
D) 110

Accepted Answer

The number of ways to select a committee of three men from 11 men is $C(11, 3) = \frac{11!}{3!(11-3)!}$, and the number of ways to select a committee of four women from 11 women is $C(11, 4) = \frac{11!}{4!(11-4)!}$. The total number of ways to form the committee is the product of these two, which is $C(11, 3) \times C(11, 4) = 165 \times 330 = 54450$.

Question 10

Compute Probabilities Involving Permutations and Combinations
-From 8 names on a ballot, a committee of 5 will be elected to attend a political national convention. How many different committees are possible
A) 56
B) 6720
C) 336
D) 3360

Accepted Answer

The answer of Compute Probabilities Involving Permutations and Combinations
-From 8...

Question 11

Solve Counting Problems Involving Permutations with Nondistinct Items
-Given the sets of digits {1, 3, 4, 6, 8}, how many different numbers between 40,000 and 80,000 can be written using these digits if repetition of digits is allowed
A) 1250
B) 3125
C) 120
D) 1875
E) 22

Accepted Answer

The answer of Solve Counting Problems Involving Permutations with Nondistinct...

Question 12

Solve Counting Problems Involving Permutations with Nondistinct Items
-How many distinct arrangements of the letters in the word football are possible
A) 10,080
B) 1680
C) 720
D) 40,320

Accepted Answer

The word "football" has 8 letters with "o" repeated twice and "l" repeated twice. The formula for arrangements with repeated letters is n! / (p1! * p2! * ... * pk!), where n is the total number of letters, and p1, p2, ..., pk are the frequencies of each repeated letter. Here, it's 8! / (2! * 2!) = 40,320 / (2 * 2) = 10,080.

Question 13

Compute Probabilities Involving Permutations and Combinations
-A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 6 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen
A) 1680
B) 972
C) 261,273,600
D) 7,257,600

Accepted Answer

The number of ways to choose 6 multiple-choice questions out of 9 is calculated using combinations: $C(9,6) = \frac{9!}{6!(9-6)!} = 84$. The number of ways to choose 3 open-ended problems out of 6 is $C(6,3) = \frac{6!}{3!(6-3)!} = 20$. The total number of ways to choose the questions and problems is the product of these two, which is $84 \times 20 = 1680$.

Question 14

Compute Probabilities Involving Permutations and Combinations
-A box contains 26 widgets, 4 of which are defective. If 4 are sold at random, find the probability that (a) all are defective (b) none are defective.
A) $\frac { 1 } { 14,950 } ; \frac { 1463 } { 2990 }$
B) $\frac { 2 } { 13 } ; \frac { 11 } { 13 }$
C) $\frac { 1 } { 358,800 } ; \frac { 1 } { 89,700 }$
D) $\frac { 1 } { 26 } ; \frac { 2 } { 13 }$

Accepted Answer

The probability that all 4 widgets sold are defective is calculated using combinations: $\frac{{\binom{4}{4}}}{\binom{26}{4}} = \frac{1}{14,950}$. The probability that none are defective is calculated by choosing 4 out of the 22 non-defective widgets: $\frac{{\binom{22}{4}}}{\binom{26}{4}} = \frac{1463}{2990}$.

Question 15

Compute Probabilities Involving Permutations and Combinations
-To play the lottery in a certain state, a person has to correctly select 5 out of 45 numbers, paying $\$ 1$ for each five-number selection. If the five numbers picked are the same as the ones drawn by the lottery, an enormous sum of money is bestowed. What is the probability that a person with one combination of five numbers will win? What is the probability of winning if 100 different lottery tickets are purchased?
A) $\frac { 1 } { 1,221,759 } ; \frac { 100 } { 1,221,759 }$
B) $\frac { 1 } { 146,611,080 } ; \frac { 10 } { 14,661,108 }$
C) $\frac { 1 } { 8,145,060 } ; \frac { 1 } { 814,506 }$
D) $\frac { 1 } { 5,864,443,200 } ; \frac { 1 } { 58,644,432 }$

Accepted Answer

The probability of winning with one ticket is calculated using combinations since the order in which the numbers are drawn does not matter. The total number of ways to choose 5 numbers out of 45 is $ \binom{45}{5} $, which equals 1,221,759. Therefore, the probability of winning with one ticket is $ \frac{1}{1,221,759} $. If 100 different tickets are purchased, assuming each ticket has a unique combination, the probability of winning increases linearly, making it $ \frac{100}{1,221,759} $.

Question 16

Compute Probabilities Involving Permutations and Combinations
-A committee consisting of 6 people is to be selected from eight parents and four teachers. Find the probability of selecting three parents and three teachers.
A) $\frac { 8 } { 33 }$
B) $\frac { 2 } { 33 }$
C) $\frac { 100 } { 231 }$
D) $\frac { 10 } { 11 }$

Accepted Answer

The probability of selecting three parents and three teachers is the number of ways to select three parents from eight parents multiplied by the number of ways to select three teachers from four teachers, divided by the total number of ways to select six people from twelve people.The number of ways to select three parents from eight parents is $\binom{8}{3} = 56$.The number of ways to select three teachers from four teachers is $\binom{4}{3} = 4$.The total number of ways to select six people from twelve people is $\binom{12}{6} = 924$.Therefore, the probability of selecting three parents and three teachers is $\frac{56 \times 4}{924} = \frac{2}{33}$.

Question 17

Compute Probabilities Involving Permutations and Combinations
-Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly and each person arrives at a different time. In how many ways can they arrive? In how many ways can Jean arrive first and Keith last? Find the probability that Jean will arrive first and Keith will arrive last.
A) $720 ; 24 ; \frac { 1 } { 30 }$
B) $720 ; 15 ; \frac { 1 } { 48 }$
C) $120 ; 6 ; \frac { 1 } { 20 }$
D) $120 ; 10 ; \frac { 1 } { 12 }$

Accepted Answer

The total number of ways they can arrive is $6!$ because there are 6 people and each has a unique position in the sequence of arrivals, which equals $720$. For Jean to arrive first and Keith last, the other 4 can arrive in any order, which is $4!$ ways, equaling $24$. The probability is the number of favorable outcomes ($24$) over the total number of outcomes ($720$), which simplifies to $\frac{1}{30}$.

Question 18

Compute Probabilities Involving Permutations and Combinations
-If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that none of the 5 cards are picture cards.
A) $\frac { 108,257 } { 108,290 }$
B) $\frac { 3 } { 13 }$
C) $\frac { 33 } { 108,290 }$
D) $\frac { 1 } { 216,580 }$

Accepted Answer

There are 12 picture cards in a deck (4 Jacks, 4 Queens, 4 Kings) and 40 non-picture cards. The probability of drawing 5 non-picture cards is the number of ways to choose 5 cards from the 40 non-picture cards divided by the number of ways to choose 5 cards from all 52 cards. This is calculated as $\frac{{\binom{40}{5}}}{{\binom{52}{5}}} = \frac{658,008}{2,598,960} = \frac{108,257}{108,290}$.

Question 19

Compute Probabilities Involving Permutations and Combinations
-To win at LOTTO in a certain state, one must correctly select 6 numbers from a collection of 55 numbers (one through 55.) The order in which the selections is made does not matter. How many different selections are possible
A) 28,989,675
B) 25,827,165
C) 330
D) 720

Accepted Answer

To calculate the number of different selections, we use the formula for combinations which is:

nCr = n! / r!(n-r)!

where n is the total number of options (in this case, 55) and r is the number of selections (in this case, 6).

Plugging in the values, we get:

55C6 = 55! / 6!(55-6)!
= 55! / 6!49!
= (55 × 54 × 53 × 52 × 51 × 50) / (6 × 5 × 4 × 3 × 2 × 1)
= 28,989,675

Therefore, the best choice is A, which is the number of different selections possible.

Question 20

Compute Probabilities Involving Permutations and Combinations
-A group consists of 6 men and 5 women. Four people are selected to attend a conference. In how many ways can 4 people be selected from this group of 11 ? In how many ways can 4 men be selected from the 6 men? Find the probability that the selected group will consist of all men.
A) $330 ; 15 ; \frac { 1 } { 22 }$
B) $7920 ; 360 ; \frac { 1 } { 22 }$
C) $330 ; 15 ; \frac { 1 } { 15,840 }$
D) $330 ; 15 ; \frac { 1 } { 1,814,400 }$

Accepted Answer

The total ways to select 4 people from 11 is $C(11, 4) = 330$. The ways to select 4 men from 6 is $C(6, 4) = 15$. The probability of selecting 4 men is $\frac{15}{330} = \frac{1}{22}$.

Quiz 5: Probability

Compute Probabilities Involving Permutations and Combinations -If you are dealt 6 cards from a shuffled deck of 52 cards, find the probability of getting 3 jacks and 3 aces.

Determine the Appropriate Probability Rule to Use -Find $P ( A$ or $B )$ given that $P ( A ) = 0.7 , P ( B ) = 0.2$ , and $A$ and $B$ are mutually exclusive.

Solve Counting Problems Involving Permutations with Nondistinct Items -How many distinct arrangements can be formed from all the letters of statistics

Compute Probabilities Involving Permutations and Combinations -A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 8 members and any member can be elected to each position No person can hold more than one office.

Solve Counting Problems Involving Permutations with Nondistinct Items -How many distinct arrangements of the letters in the word Mississippi are possible

Solve Counting Problems Involving Permutations with Nondistinct Items -A man has 12 coins that consist of 3 pennies, 4 nickels, and 5 quarters. How many distinct arrangements of the coins can he make if he lays them in a row one at a time

Compute Probabilities Involving Permutations and Combinations -If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that all 5 cards are picture cards.

Compute Probabilities Involving Permutations and Combinations -In how many ways can a committee of three men and four women be formed from a group of 11 men and 11 women

Compute Probabilities Involving Permutations and Combinations -From 8 names on a ballot, a committee of 5 will be elected to attend a political national convention. How many different committees are possible

Solve Counting Problems Involving Permutations with Nondistinct Items -Given the sets of digits {1, 3, 4, 6, 8}, how many different numbers between 40,000 and 80,000 can be written using these digits if repetition of digits is allowed

Solve Counting Problems Involving Permutations with Nondistinct Items -How many distinct arrangements of the letters in the word football are possible

Compute Probabilities Involving Permutations and Combinations -A box contains 26 widgets, 4 of which are defective. If 4 are sold at random, find the probability that (a) all are defective (b) none are defective.

Compute Probabilities Involving Permutations and Combinations -A committee consisting of 6 people is to be selected from eight parents and four teachers. Find the probability of selecting three parents and three teachers.

Compute Probabilities Involving Permutations and Combinations -If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that none of the 5 cards are picture cards.

Compute Probabilities Involving Permutations and Combinations -To win at LOTTO in a certain state, one must correctly select 6 numbers from a collection of 55 numbers (one through 55.) The order in which the selections is made does not matter. How many different selections are possible

Quiz 5: Probability

Compute Probabilities Involving Permutations and Combinations -If you are dealt 6 cards from a shuffled deck of 52 cards, find the probability of getting 3 jacks and 3 aces.

Determine the Appropriate Probability Rule to Use -Find P(AP ( AP(A or B) B ) B) given that P(A) =0.7,P(B) =0.2P ( A ) = 0.7 , P ( B ) = 0.2P(A) =0.7,P(B) =0.2 , and AAA and BBB are mutually exclusive.

Solve Counting Problems Involving Permutations with Nondistinct Items -How many distinct arrangements can be formed from all the letters of statistics

Compute Probabilities Involving Permutations and Combinations -A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 8 members and any member can be elected to each position No person can hold more than one office.

Solve Counting Problems Involving Permutations with Nondistinct Items -How many distinct arrangements of the letters in the word Mississippi are possible

Solve Counting Problems Involving Permutations with Nondistinct Items -A man has 12 coins that consist of 3 pennies, 4 nickels, and 5 quarters. How many distinct arrangements of the coins can he make if he lays them in a row one at a time

Compute Probabilities Involving Permutations and Combinations -If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that all 5 cards are picture cards.

Compute Probabilities Involving Permutations and Combinations -In how many ways can a committee of three men and four women be formed from a group of 11 men and 11 women

Compute Probabilities Involving Permutations and Combinations -From 8 names on a ballot, a committee of 5 will be elected to attend a political national convention. How many different committees are possible

Solve Counting Problems Involving Permutations with Nondistinct Items -Given the sets of digits {1, 3, 4, 6, 8}, how many different numbers between 40,000 and 80,000 can be written using these digits if repetition of digits is allowed

Solve Counting Problems Involving Permutations with Nondistinct Items -How many distinct arrangements of the letters in the word football are possible

Compute Probabilities Involving Permutations and Combinations -A box contains 26 widgets, 4 of which are defective. If 4 are sold at random, find the probability that (a) all are defective (b) none are defective.

Compute Probabilities Involving Permutations and Combinations -A committee consisting of 6 people is to be selected from eight parents and four teachers. Find the probability of selecting three parents and three teachers.

Compute Probabilities Involving Permutations and Combinations -If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that none of the 5 cards are picture cards.

Compute Probabilities Involving Permutations and Combinations -To win at LOTTO in a certain state, one must correctly select 6 numbers from a collection of 55 numbers (one through 55.) The order in which the selections is made does not matter. How many different selections are possible

Determine the Appropriate Probability Rule to Use -Find $P ( A$ or $B )$ given that $P ( A ) = 0.7 , P ( B ) = 0.2$ , and $A$ and $B$ are mutually exclusive.