# Permutation and Combination Word Problems

Permutation and Combination Word Problems
• Page 1
1.
Find the probability of getting exactly 2 heads when 5 coins are tossed simultaneously.
 a. $\frac{1}{32}$ b. $\frac{5}{16}$ c. $\frac{5}{32}$ d. $\frac{1}{4}$

#### Solution:

Probability of getting r heads from n trials = nCr · pr · qn - r
[Formula for the probability of a binomial distribution.]

Here n = 5; p = 1 / 2; q = 1 / 2; r = 2

Probability of getting exactly 2 heads = 5C2 · (1 / 2)2 · (1 / 2)3 = 10 / 32 = 5 / 16

2.
If on an average it rains on 5 days in every thirty days, then find the probability that it rains exactly on 3 days of a given week.
 a. 7C3 · ${\left(\frac{5}{6}\right)}^{7}$ b. 7C3 · ${\left(\frac{5}{6}\right)}^{3}$ · ${\left(\frac{1}{6}\right)}^{4}$ c. 7C3 · ${\left(\frac{1}{6}\right)}^{3}$ · ${\left(\frac{5}{6}\right)}^{4}$ d. 7C3 · ${\left(\frac{1}{6}\right)}^{7}$

#### Solution:

The probability that it rains in a day = p = 5 / 30 = 1 / 6

The probability that it does not rain = q = 1 - p = 5 / 6
[p + q = 1.]

Considering that the probabilities of outcomes follow the binomial distribution and setting n = 7 and r = 3 in the formula P(r) = nCr · pr · qn - r

Probability that it rains 3 days in a week = 7C3 · (1 / 6)3 · (5 / 6)4

3.
The probability that a bulb manufactured by a factory will fuse before 100 days of use is 0.05. Find the probability that out of 5 such bulbs at least one will fuse before 100 days of use.
 a. 5 · ($\frac{19}{20}$)5 b. 1 - ($\frac{19}{20}$)5 c. 1 - ($\frac{1}{20}$)5 d. 5 · ($\frac{1}{20}$)5

#### Solution:

The probability that a bulb fuses before 100 days of use = p = 0.05 = 1 / 20

The probability that a bulb does not fuse before100 days of use = q = 1 - p = 0.95 = 19 / 20

The probability that out of 5 bulbs at least one will fuse before 100 days of use = 1 - the probability that none of the bulbs will fuse.

Considering that the probabilities of outcomes follow the binomial distribution and setting n = 5 and r = 0 in the formula P(r) = nCr · pr · qn - r

The probability that all 5 bulbs will not fuse = 5C0 (1 / 20)0 (19 / 20)5 = (19 / 20)5

So, the probability that out of 5 bulbs at least one will fuse before 100 days of use = 1 - (19 / 20)5.

4.
An insurance agent sells 5 policies, one to each of 5 healthy persons of the same age. According to the actuarial tables the probability that a man of this particular age will live for 35 years hence is $\frac{2}{3}$. Find the probability that after 35 years, at least 3 men will be alive.
 a. $\frac{192}{243}$ b. $\frac{32}{243}$ c. $\frac{131}{243}$ d. $\frac{242}{243}$

#### Solution:

The probability that a person is alive 35 years hence is p = 2 / 3

The probability that a person is not alive 35 years hence is q = 1 - (2 / 3) = 1 / 3

The probability that at least 3 men will be alive = the probability that exactly 3 will be alive + the probability that exactly 4 will be alive + the probability that exactly 5 will be alive

Considering that the probabilities of outcomes follow the binomial distribution P(X = r) = nCr · pr · qn - r and setting n = 5 and giving values for r = 3, 4, 5.

P (r = 3) + P (r = 4) + P (r = 5) = 5C3 · (2 / 3)3 · (1 / 3)2 + 5C4 · (2 / 3)4 · (1 / 3) + 5C5 · (2 / 3)5

= [(10×8)+(5×16)+32]243
[Simplify.]

= 192 / 243

The probability that at least 3 men will be alive = 192 / 243

5.
An insurance agent sells 5 policies, one to each of 5 persons who are of the same age and in good health. According to the actuarial tables the probability that a man of this particular age will be alive 35 years hence is 2/3. Find the probability that after 35 years, at most 3 men will be alive.
 a. $\frac{32}{243}$ b. $\frac{242}{243}$ c. $\frac{192}{243}$ d. $\frac{131}{243}$

#### Solution:

The probability that a person is alive 35 years hence is p = 2 / 3

The probability that a person is not alive 35 years hence is q = 1 - (2 / 3) = 1 / 3

Considering that the probabilities of outcomes follow the binomial distribution P(X = r) = nCr · pr · qn - r and setting n = 5 and giving values for r = 4, 5.

The probability that at most 3 men will be alive = 1 - [the probability that exactly 4 will be alive + the probability that all the 5 will be alive]

= 1 - [5C4 · (2 / 3)4 · (1 / 3) + 5C5 · (2 / 3)5]

= 1 - [(80+32)243]
[Simplify.]

= 131 / 243
[Simplify.]

The probability that at most 3 will be alive is 131 / 243 .

6.
If on an average one in every 10 trains arrives late, then find the probability that out of 5 trains expected to arrive, exactly 4 will arrive in time.
 a. 5C4 · ${\left(\frac{1}{10}\right)}^{4}$ · ($\frac{9}{10}$) b. 5C4 · ${\left(\frac{9}{10}\right)}^{4}$ · ($\frac{1}{10}$) c. 5C4 · ${\left(\frac{1}{10}\right)}^{4}$ d. 5C4 · ${\left(\frac{9}{10}\right)}^{4}$

#### Solution:

Probability for a train to be late is q = 1 / 10

Probability for a train to arrive in time is p = 1 - q = 9 / 10

This is a binomial situation where n = 5, r = 4, P(X = r) = nCr · pr · qn - r.

Probability that exactly 4 trains out of 5 arrive in time = 5C4 · (910)4 · (1 / 10)

7.
Find the probability of guessing correct answers to at least 9 out of 10 questions with 4 choices of answers.
 a. $\frac{31}{{\left(4\right)}^{10}}$ b. 10 × $\frac{{3}^{9}}{{4}^{10}}$ c. 10 × 39 + $\frac{10}{{4}^{10}}$ d. ($\frac{10}{4}$)10

#### Solution:

Probability that an answer to a question is correct is p = 1 / 4
[One out of 4 choices is correct.]

Probability that an answer to a question is wrong q is 1 - p = 3 / 4
[Sum of probabilities = 1.]

Probability of guessing atleast 9 out of 10 questions correctly = probability of guessing 9 questions correctly + probability of guessing 10 questions correctly.

P(at least 9) = P(r = 9) + P(r = 10)

This is a binomial situation and substituting n = 10 and r = 9, 10 in the formula.

P(X = r) = nCr · pr · qn - r

P(at least 9) = P(r = 9) + P(r = 10) = 10C9 · (1 / 4)9 · (3 / 4) + 10C10 · (1 / 4)10 = 31(4)10
[Simplify.]

So, the probability of guessing correctly to at least 9 out of 10 questions is 31(4)10.

8.
A survey found that three out of every five students play basketball. If 10 students are selected at random, then find the probability that exactly 5 will play basketball.
 a. 10C5 b. 0.5 c. 0.6 d. 0.20

#### Solution:

In a binomial experiment, the probability of exactly r successes in n trials is P(X = r) = n!(n-r)!r! . pr . qn-r

Probability that a student selected plays basketball is p = 3 / 5

Probability that a student selected does not play basketball is q = 1 - p = 2 / 5

n = 10, r = 5

P(X = 5) = 10!(10-5)!5!. (3 / 5) 5. (2 / 5) 10-5 = 0.20
[Substitute the values of p, q, n, r.]

The probability that exactly 5 students will play basketball is 0.20.

9.
In a survey it was found that 20% of people living in a society watch channel A. If a random sample of 10 people is selected, then find the probability that exactly 5 people in the sample watch channel A.
 a. 0.5 b. 0.2 c. 0.026 d. 0.26

#### Solution:

In a binomial experiment, the probability of exactly r successes in n trials is P(X = r) = n!(n-r)!r! . pr . qn-r

Probability that a person selected watches channel 'A' is p = 20% = 0.2

Probability that a person selected does not watch channel 'A' is q = 1 - p = 0.8

n = 10, r = 5

P(X = 5) = 10!(10-5)!5! (0.2 ) 5 (0.8 ) 10-5 = 0.026
[Substitute the values of p, q, n, r.]

The probability that exactly 5 people in the sample watch channel A is 0.026.

10.
In a survey it was found that 20% of people living in a society watch channel A. If a random sample of 10 people are selected, then find the probability that at most 3 people in the sample watch channel A.
 a. 0.878 b. 0.2 c. 0.79 d. 0.59

#### Solution:

In a binomial experiment, the probability of exactly r successes in n trials is P(X = r) = n!(n-r)!r! . pr . qn-r

Probability that a person selected watches channel 'A' is p = 20% = 0.2

Probability that a person selected does not watch channel 'A' is q = 1 - p = 0.8

n = 10

"At most 3 people" means 0, or 1, or 2, or 3. Hence r = 0, 1, 2, 3.

P(at most 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = 0) = 10!(10-0)!0! (0.2 ) 0 (0.8 ) 10-0 = 0.107
[Simplify.]

P(X = 1) = 10!(10-1)!1! (0.2 ) 1 (0.8 ) 10-1 = 0.27
[Simplify.]

P(X = 2) = 10!(10-2)!2! (0.2 ) 2 (0.8 ) 10-2 = 0.3
[Simplify.]

P(X = 3) = 10!(10-3)!3! (0.2 ) 3 (0.8 ) 10-3 = 0.201
[Simplify.]

P(at most 3) = 0.107 + 0.27 + 0.3 + 0.2 = 0.878
[Substitute and simplify.]

The probability that at most 3 people will watch channel A is 0.878.