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. | $\frac{1}{4}$ |

[Formula for the probability of a binomial distribution.]

Here

Probability of getting exactly 2 heads = 5

Correct answer : (2)

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. | 7 ^{C}3 · ${(\frac{5}{6})}^{7}$ | ||

b. | 7 ^{C}3 · ${(\frac{5}{6})}^{3}$ · ${(\frac{1}{6})}^{4}$ | ||

c. | 7 ^{C}3 · ${(\frac{1}{6})}^{3}$ · ${(\frac{5}{6})}^{4}$ | ||

d. | 7 ^{C}3 · ${(\frac{1}{6})}^{7}$ |

The probability that it does not rain =

[

Considering that the probabilities of outcomes follow the binomial distribution and setting

Probability that it rains 3 days in a week = 7

Correct answer : (3)

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} |

The probability that a bulb does not fuse before100 days of use =

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

The probability that all 5 bulbs will not fuse = 5

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

Correct answer : (2)

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. | $\frac{242}{243}$ |

The probability that a person is not alive 35 years hence is

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 =

P (

=

[Simplify.]

=

The probability that at least 3 men will be alive =

Correct answer : (1)

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. | $\frac{131}{243}$ |

The probability that a person is not alive 35 years hence is

Considering that the probabilities of outcomes follow the binomial distribution P(X =

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 - [5

= 1 - [

[Simplify.]

=

[Simplify.]

The probability that at most 3 will be alive is

Correct answer : (4)

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. | 5 ^{C}4 · ${(\frac{1}{10})}^{4}$ · ($\frac{9}{10}$) | ||

b. | 5 ^{C}4 · ${(\frac{9}{10})}^{4}$ · ($\frac{1}{10}$) | ||

c. | 5 ^{C}4 · ${(\frac{1}{10})}^{4}$ | ||

d. | 5 ^{C}4 · ${(\frac{9}{10})}^{4}$ |

Probability for a train to arrive in time is

This is a binomial situation where

Probability that exactly 4 trains out of 5 arrive in time = 5

Correct answer : (2)

7.

Find the probability of guessing correct answers to at least 9 out of 10 questions with 4 choices of answers.

a. | $\frac{31}{{(4)}^{10}}$ | ||

b. | 10 × $\frac{{3}^{9}}{{4}^{10}}$ | ||

c. | 10 × 3 ^{9} + $\frac{10}{{4}^{10}}$ | ||

d. | ($\frac{10}{4}$) ^{10} |

[One out of 4 choices is correct.]

Probability that an answer to a question is wrong

[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

P(X =

P(at least 9) = P(

[Simplify.]

So, the probability of guessing correctly to at least 9 out of 10 questions is

Correct answer : (1)

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. | _{10}C_{5} | ||

b. | 0.5 | ||

c. | 0.6 | ||

d. | 0.20 |

Probability that a student selected plays basketball is

Probability that a student selected does not play basketball is

P(X = 5) =

[Substitute the values of

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

Correct answer : (4)

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 |

Probability that a person selected watches channel 'A' is

Probability that a person selected does not watch channel 'A' is

P(X = 5) =

[Substitute the values of

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

Correct answer : (3)

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 |

Probability that a person selected watches channel 'A' is

Probability that a person selected does not watch channel 'A' is

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

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

P(X = 0) =

[Simplify.]

P(X = 1) =

[Simplify.]

P(X = 2) =

[Simplify.]

P(X = 3) =

[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.

Correct answer : (1)