﻿ Independent Events Worksheets | Problems & Solutions

# Independent Events Worksheets

Independent Events Worksheets
• Page 1
1.
A box consists cards labeled 1 through 10. A card is drawn at random and replaced. Then a second card is drawn. Find the probability that the first number is odd and the second number is a multiple of 3.
 a. 0.08 b. 1.5 c. 0.15 d. 0.8

#### Solution:

P (1st number is odd) = 510 = 1 / 2
[There are 5 odd numbers: 1, 3, 5, 7, 9.]

P (2nd number is multiple of 3) = 310
[There are 3 multiples of three: 3, 6, 9.]

P (1st number is odd and 2nd number is multiple of 3)

= P (1st number is odd) × P (2nd number is multiple of 3)

= 12 × 310

= 320 = 0.15

= 320 = 0.15

The probability that the first number is odd and the second number is a multiple of 3 is 0.15.

2.
A box consists cards labeled 1 through 10 written on them. A card is drawn at random and replaced. Then a second card is drawn. Find the probability that the first number is less than 5 and second number is a prime number.
 a. 1.6 b. 0.08 c. 0.8 d. 0.16

#### Solution:

P (1st number less than 5) = 410
[There are 4 numbers that are less than 5 : 1, 2, 3, 4]

P (2nd number is prime) = 410
[There are 4 prime numbers : 2, 3, 5, 7.]

P (1st number is less than 5 and 2nd number is prime) = P (1st number is less than 5) × P (2nd number is prime)

= 410 × 410

= 0.16

The probability that the first number is less than 5 and second number is a prime number is 0.16.

3.
The probability that A gets a fellowship is 0.3 and B gets fellowship is 0.8. Find the probability that both A and B get fellowship.
 a. 0.24 b. 0.14 c. 0.11 d. 2.4

#### Solution:

P(A gets fellowship) = 0.3

P(B gets fellowship) = 0.8

P(A and B get fellowship) = P(A gets fellowship) × P(B gets fellowship)
[A and B are independent events.]

= 0.3 × 0.8 = 0.24

4.
If P(A) = 0.1, P(B) = 0.4 and P(A$\cap$B) = 0.2, then find P(A$\cup$B)
 a. 0.3 b. 0.7 c. 0.5 d. 0.6

#### Solution:

P (AB) = 0.1 + 0.4 Ã¢â‚¬â€œ 0.2
[P (AB) = P (A) + P (B) - P (AB).]

= 0.3

5.
If S is a sample space and A, B and C are mutually exclusive, then P(A$\cup$B$\cup$C) = ?
 a. 1 b. P (A) + P (B) + P(C) - P(A$\cap$B$\cap$C) c. P (A) + P (B) + P(C)

#### Solution:

P (ABC) = P (A) + P (B) + P(C) Ã¢â‚¬â€œ P (AB) - P (BC) - P(CA) + P (ABC)

= P (A) + P (B) + P(C).
[ A, B and C are mutually exclusive.
P (AB) = 0; P (BC) = 0; P (CA) = 0 and P (ABC) = 0.]

6.
If P (A) = 0.6, then P ($\stackrel{‾}{A}$) = ?
 a. 0.4 b. 0.6 c. 1

#### Solution:

P (A) = 1 - P (A)

= 1 - 0.6

= 0.4

7.
If P ($\stackrel{‾}{E}$) = 0.02, then P (E) = ?
 a. 0.08 b. 0.98 c. 0.02 d. 0.8

#### Solution:

P (E) = 1 - P (E)

= 1 - 0.02

= 0.98

8.
If P(A) = 0.3, P(A$\cap$B) = 0.06 and the events A and B are independent, then find P(B)?
 a. 0.2 b. 0.02 c. 0.24 d. 0.03

#### Solution:

P(AB) = P(A) × P(B)
[A and B are independent events.]

0.06 = 0.3 × P(B)

0.2 = P(B)

9.
If P (A) = 0.25 and P (B) = 0.5, then find P(A$\cap$B), if A and B are independent events.
 a. 1.25 b. 0.3 c. 0.75 d. 0.125

#### Solution:

P(AB) = 0.25 x 0.5
[P(AB) = P(A) x P(B).]

= 0.125

10.
A box consists cards labeled numbers 1 through 10. A card is drawn at random and replaced. Then a second card is drawn. Find the probability that the first number drawn is 5 and the second number is 10.
 a. 0.1 b. 0.01 c. 1 d. 0.2

#### Solution:

Let A be the event of getting a number 5.

P(A) = 110

Let B be the event of getting a number 10.

P(B) = 110

P(the first number is 5 and the second number is 10) = P(A B)

= P(A) × P(B)
[A and B are independent events.]

= 110 × 110

= 1100 = 0.01

The probability that the first number drawn is 5 and the second number is 10 is 0.01.