Probability Problems

**Page 1**

1.

Find the probability of getting a sum 7 from 3 throws of a dice?

a. | $\frac{1}{36}$ | ||

b. | $\frac{1}{16}$ | ||

c. | $\frac{1}{9}$ | ||

d. | $\frac{1}{6}$ |

In 3 throws of a dice n(s) = 6$\times $6$\times $6 = 216

Let E = getting a sum = {(3,4),(5,2),(6,1),(4,3),(2,5),(1,6)}

P(E) = $\frac{n(E)}{n(s)}$

P(E) = $\frac{6}{216}$

P(E) = $\frac{1}{36}$

Answer :

$\frac{1}{36}$

Correct answer : (1)

2.

In a bag you have 11 red,4 blue and 9 orange balls. Suppose you picked up 1 ball randomly. Find the probability that it is neither red nor orange?

a. | $\frac{1}{12}$ | ||

b. | $\frac{1}{2}$ | ||

c. | $\frac{1}{6}$ | ||

d. | $\frac{1}{3}$ |

Given: total number of balls = 11 + 4 + 9 = 24

n(E) = 4 [blue ball drawn]

P(E) = $\frac{n(E)}{n(s)}$

P(E) = $\frac{4}{24}$

P(E) = $\frac{1}{6}$

Answer:

Probability that it is neither red nor orange = $\frac{1}{6}$

Correct answer : (3)

3.

In a bag there are numbered 1 to 20 tickets then 1 ticket is drawn randomly. Find the probability that ticket drawn has a number which is a multiple of 5 and 7?

a. | $\frac{1}{3}$ | ||

b. | $\frac{3}{20}$ | ||

c. | $\frac{3}{5}$ | ||

d. | $\frac{3}{10}$ |

Given : numbers S = {1,2,3,4,.......,18,19,20}

Let E = event of getting a multiple of 5 and 7 = {5,7,10,14,15,20}

P(E) = $\frac{n(E)}{n(S)}$

P(E) = $\frac{6}{20}$

P(E) = $\frac{3}{10}$

Answer:

The probability that ticket drawn has a number which is a multiple of 5 and 7 = $\frac{3}{10}$

Correct answer : (4)

4.

Find the probability of getting a face card from an ordinary deck of 52 playing cards?

a. | $\frac{1}{13}$ | ||

b. | $\frac{3}{13}$ | ||

c. | $\frac{4}{13}$ | ||

d. | $\frac{2}{13}$ |

Given : playing cards n(S) = 52

and n(E)= event of getting a face card = 12

P(E) = $\frac{n(E)}{n(S)}$

P(E) = $\frac{12}{52}$

P(E) = $\frac{3}{13}$

Answer:

The probability of getting a face card = $\frac{3}{13}$

Correct answer : (2)

5.

Find the probability of getting a non-face card or a 7 card from an ordinary deck of 52 playing cards?

a. | $\frac{10}{13}$ | ||

b. | $\frac{1}{13}$ | ||

c. | $\frac{3}{13}$ | ||

d. | $\frac{7}{13}$ |

Given : playing cards n(S) = 52

and n(non-face)= event of getting a non-face card = 40

n(7)=event of getting a 7 card = 4

n(non-face and 7) = 4/52

P(E) = $\frac{n(E)}{n(S)}$

P(non-face) = $\frac{40}{52}$

P(7) = $\frac{4}{52}$

P(non-face and 7) = $\frac{4}{52}$

P(non-face or 7) = P(non-face)+P(7)-P(non-face and 7)

P(non-face or 7) = $\frac{40}{52}$+ $\frac{4}{52}$- $\frac{4}{52}$

P(non-face or 7) = $\frac{40}{52}$

P(non-face or 7) = $\frac{10}{13}$

Answer :

The probability of getting a non-face card or a 7 card = $\frac{10}{13}$

Correct answer : (1)

6.

Tom tossed 2 coins. Find the probability of getting two heads?

a. | $\frac{1}{3}$ | ||

b. | $\frac{1}{2}$ | ||

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

d. | $\frac{1}{6}$ |

As we know each coin has two possible outcomes Head and tail

Here we have 2 coins

So, n(S) = {(H,T)(T,H)(H,H)(T,T)}

Getting two heads n(E)={(H,H)}

P(E) = $\frac{n(E)}{n(S)}$

P(E) = $\frac{1}{4}$

Answer:

The probability of getting heads = $\frac{1}{4}$

Correct answer : (3)

7.

A family has 3 children, if the probability of giving birth to a boy is same as the giving birth to a girl. Find the probability getting at least two boys?

a. | $\frac{1}{3}$ | ||

b. | $\frac{1}{2}$ | ||

c. | $\frac{1}{8}$ | ||

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

Given : 3 children

Children in order n(S) = {(BBB),(BBG),(BGB),(BGG),(GGG),(GGB),(GBG)(GBB)} = 8

Event of getting 2 boys n(E) = {(BBB),(BBG),(BGB),(GBB)} = 4

P(E) = $\frac{n(E)}{n(S)}$

P(E) = $\frac{4)}{8}$

P(E) = $\frac{1}{2}$

Answer:

The probability of getting at least 2 boys = P(E) = $\frac{1}{2}$

Correct answer : (2)

8.

Find the probability of getting two numbers whose product is odd. If two dice are thrown simultaneously.

a. | $\frac{2)}{7}$ | ||

b. | $\frac{1)}{10}$ | ||

c. | $\frac{1)}{8}$ | ||

d. | $\frac{2)}{9}$ |

Two dice thrown simultaneously, n(S) = 6$\times$ 6 = 36

Event of getting two numbers product odd n(E) = {(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)} = 8

P(E)=$\frac{n(E)}{n(S)}$

P(E)=$\frac{8}{36}$

P(E)=$\frac{2}{9}$

Answer:

The probability of getting two numbers whose product is odd = $\frac{2)}{9}$

Correct answer : (4)

9.

In a box, there are 7 marbles, 3 black and 4 white. What is the probability that a black marble drawn?

a. | $\frac{1}{7}$ | ||

b. | $\frac{2}{7}$ | ||

c. | $\frac{4}{7}$ | ||

d. | $\frac{3}{7}$ |

Given marbles n(S) = 7

Black marble n(E) =3

P(E) = $\frac{n(E)}{n(S)}$

P(E) = $\frac{3}{7}$

Answer:

The probability of getting a black marble = $\frac{3}{7}$

Correct answer : (4)

10.

In a room there are 10 girls and 15 boys. Find the probability of selecting at least 2 girls and 1 boy, if 3 children are selected randomly?

a. | $\frac{23}{90}$ | ||

b. | $\frac{27}{92}$ | ||

c. | $\frac{29}{95}$ | ||

d. | $\frac{24}{93}$ |

Let S = sample space

E = event of selecting 2 girls and 1 boy

n(S) = $^{25}C_{3}$

n(S) = 2300

n1(E) = $^{20}C_{2}$ $\Times$$^{15}C_{1}$

n(E) = 675

Then,

P(E) = $\frac{n(E)}{n(S)}$

P(E) = $\frac{675}{2300}$

P(E) = $\frac{27}{92}$

Answer:

The probability of selecting at least 2 girls and 1 boy = $\frac{27}{92}$

Correct answer : (2)