Question 1

Suppose that the sample space is S = {a, b, c, d, e, f, g, h} and that outcomes are equally likely. Find theprobability of the event E = {b, f, h}.

Accepted Answer

The probability of an event E is the number of outcomes in E divided by the total number of outcomes in the sample space S. Here, |E| = 3 (b, f, h) and |S| = 8 (a, b, c, d, e, f, g, h), so the probability of E is 3/8 = 0.375.

Question 2

A box contains 20 widgets, 4 of which are defective. If 4 are sold at random, find the probability that (a) all aredefective (b) none are defective.

Accepted Answer

The probability of all 4 widgets being defective is calculated using combinations: $\frac{{\binom{4}{4}\binom{16}{0}}}{{\binom{20}{4}}} = \frac{1}{4845}$. The probability of none being defective is $\frac{{\binom{4}{0}\binom{16}{4}}}{{\binom{20}{4}}} = \frac{364}{969}$.

Question 3

The age distribution of members of a gymnastics association is shown in the table. A member of theassociation is selected at random. Find the probability that the person selected is between 26 and 35 inclusive.Round your answer to three decimal places. $$\begin{array} { l c } 
\text { Age (years) } & \text { Frequency } \
\hline \text { Under 21 } & 402 \
21 - 25 & 417 \
26 - 30 & 201 \
31 - 35 & 50 \
\text { Over } 35 & 24
\end{array}$$ 1094

Accepted Answer

The probability is calculated by dividing the number of members aged 26-35 (201 + 50 = 251) by the total number of members (1094). So, 251/1094 = 0.229.

Question 4

Suppose that the sample space is S = {a, b, c, d, e, f, g, h, i, j} and that outcomes are equally likely. Find theprobability of the event F = ʺa vowelʺ.

Accepted Answer

The vowels in the sample space S = {a, e, i} are 3 out of 10 total outcomes, so the probability is 3/10 = 0.3.

Question 5

A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible ifthere are 15 members and any member can be elected to each position? No person can hold more than oneoffice.

Accepted Answer

The number of ways to choose a president, vice-president, and secretary-treasurer from 15 members, with no repetition, is a permutation of 15 items taken 3 at a time: 15P3 = 15! / (15-3)! = 15 * 14 * 13 = 2730.

Question 6

Suppose that S and T are two events, P(S) = 0.59 and P(T S) = 0.16. What is P(S and T)?

Accepted Answer

P(S and T) = P(S) * P(T|S) = 0.59 * 0.16 = 0.0944.

Question 7

If you are dealt 6 cards from a shuffled deck of 52 cards, find the probability of getting 3 jacks and 3 aces.

Accepted Answer

The probability of getting 3 jacks and 3 aces from a 6-card hand can be calculated using combinations. There are 4 jacks and 4 aces in a deck of 52 cards. The number of ways to choose 3 jacks out of 4 is $\binom{4}{3}$, and the number of ways to choose 3 aces out of 4 is also $\binom{4}{3}$. The total number of ways to choose 6 cards out of 52 is $\binom{52}{6}$. Therefore, the probability is $\frac{\binom{4}{3} \times \binom{4}{3}}{\binom{52}{6}} = \frac{2}{2,544,815}$.

Question 8

If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that all 5 cards are picture cards.

Accepted Answer

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

Question 9

A committee consisting of 6 people is to be selected from eight parents and four teachers. Find the probabilityof selecting three parents and three teachers.

Accepted Answer

The probability of selecting three parents from eight parents is $\frac { 8 } { 12 }\times \frac { 7 } { 11 }\times \frac { 6 } { 10 }$ = $\frac { 336 } { 1320 }$The probability of selecting three teachers from four teachers is $\frac { 4 } { 12 }\times \frac { 3 } { 11 }\times \frac { 2 } { 10 }$ = $\frac { 24 } { 1320 }$Therefore, the probability of selecting three parents and three teachers is $\frac { 336 } { 1320 }\times \frac { 24 } { 1320 }$ = $\frac { 8064 } { 1742400 }$ = $\frac { 8 } { 33 }$

Question 10

To win at LOTTO in a certain state, one must correctly select 6 numbers from a collection of 55 numbers (onethrough 55.) The order in which the selections is made does not matter. How many different selections arepossible?

Accepted Answer

There are 55 choices for the first number, 54 for the second, 53 for the third, and so on, down to 50 for the sixth. So, the number of different selections is 55 * 54 * 53 * 52 * 51 * 50 = 28,989,675.

Question 11

In one town, 51% of adults are employed in the tourism industry. What is the probability that five adultsselected at random from the town are all employed in the tourism industry?

Accepted Answer

The answer of In one town, 51% of adults are...

Question 12

In how many ways can a committee of three men and four women be formed from a group of 9 men and 9women?

Accepted Answer

The number of ways to choose 3 men out of 9 is $C(9,3) = \frac{9!}{3!(9-3)!} = 84$, and the number of ways to choose 4 women out of 9 is $C(9,4) = \frac{9!}{4!(9-4)!} = 126$. The total number of ways to form the committee is the product of these two, which is $84 \times 126 = 10584$.

Question 13

If you are dealt 5 cards from a shuffled deck of 52 cards, find the probability that none of the 5 cards are picturecards.

Accepted Answer

There are 12 picture cards in a deck (Jacks, Queens, Kings of each suit), leaving 40 non-picture cards. The probability of drawing 5 non-picture cards is the combination of choosing 5 from 40 divided by the combination of choosing 5 from 52, which is $\frac{{40 \choose 5}}{{52 \choose 5}} = \frac { 108,257 } { 108,290 }$.

Question 14

From 9 names on a ballot, a committee of 4 will be elected to attend a political national convention. How manydifferent committees are possible?

Accepted Answer

The number of different committees possible can be calculated using the combination formula, which is $C(n, k) = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items (in this case, names), and $k$ is the number of items to choose (in this case, committee members). So, $C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = 126$.

Question 15

A study conducted at a certain college shows that 54% of the schoolʹs graduates move to a different state aftergraduating. Find the probability that among 7 randomly selected graduates, at least one moves to a differentstate after graduating.

Accepted Answer

The probability that at least one graduate moves to a different state is 1 minus the probability that none of them move. The probability that a single graduate does not move is 1 - 0.54 = 0.46. For 7 graduates, the probability that none move is (0.46)^7. Therefore, the probability that at least one moves is 1 - (0.46)^7 = 0.996.

Question 16

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must beshown. If an examinee must answer 6 of the multiple-choice questions and 4 of the open-ended problems, inhow many ways can the questions and problems be chosen?

Accepted Answer

The number of ways to choose 6 multiple-choice questions from 9 is given by the combination formula C(9,6) = 84. The number of ways to choose 4 open-ended problems from 6 is C(6,4) = 15. The total number of ways to choose the questions and problems is the product of these two combinations: 84 * 15 = 1260.

Question 17

The age distribution of members of a gymnastics association is shown in the table. A member of theassociation is selected at random. Find the probability that the person selected is at least 31. Round your answerto three decimal places. $$\begin{array} { l c } 
\text { Age (years) } & \text { Frequency } \
\hline \text { Under 21 } & 407 \
21 - 25 & 401 \
26 - 30 & 214 \
31 - 35 & 54 \
\text { Over } 35 & 28 \
\hline
\end{array}$$

Accepted Answer

To find the probability that a randomly selected member is at least 31, we add the frequencies for the age groups 31 - 35 and Over 35, then divide by the total number of members. The total number of members is $407 + 401 + 214 + 54 + 28 = 1104$. The number of members at least 31 years old is $54 + 28 = 82$. Thus, the probability is $82 / 1104 \approx 0.074$.

Question 18

To play the lottery in a certain state, a person has to correctly select 5 out of 45 numbers, paying $1 for eachfive-number selection. If the five numbers picked are the same as the ones drawn by the lottery, an enormoussum of money is bestowed. What is the probability that a person with one combination of five numbers willwin? What is the probability of winning if 100 different lottery tickets are purchased?

Accepted Answer

The probability of winning with one combination is calculated by finding the number of ways to choose 5 numbers out of 45, which is $\binom{45}{5} = 1,221,759$, so the probability is $\frac{1}{1,221,759}$. If 100 different lottery tickets are purchased, the probability of winning increases linearly, making it $\frac{100}{1,221,759}$.

Question 19

Suppose that M and N are two events, P(M) = 0.20, P(N) = 0.05, and P(M and N) = 0.02. What is P(M N)?

Accepted Answer

The answer of Suppose that M and N are two...

Question 20

Of 1454 people who came into a blood bank to give blood, 357 were ineligible to give blood. Estimate theprobability that the next person who comes in to give blood will be ineligible to give blood.

Accepted Answer

The probability is estimated by dividing the number of ineligible people by the total number of people: 357/1454 ≈ 0.246.

Quiz 5: Probability