Problems on Mutually Exclusive

Type 1. (b) Mutually Exclusive

P(AUB) = P(A) + P(B) (or) P(A + B) = P(A) + P(B)

Example 1.1.12

One card is drawn from a pack of 52 cards. What is the probability that it is either a king or a queen.

Solution :

A = {an event that the card drawn is king}

P(A) = n(A)/n(S) = 4/52 = 1/13

B = {an event that the card drawn is queen}

P(B) = n(B)/n(S) = 4/52 = 1/13

AUB = {an event that the card to be either a king or a queen}

P(AUB) = P(A) + P(B) ['.' A and B are mutually exclusive events]

= (1/13) + (1/13) = 2/13

Example 1.1.13

From a group of 5 first year, 4 second year and 4 third year students, 3 students are selected at random. Find the probability that they are first year or third year students.

Solution:

Total number of students in the group = 13

Three students are selected at random n(S) = 13C3= 13.12.11 / 1.2.3 = 286

A -> The three students are from first year

i.e., n(A) = 5C₃ = 5.4.3 / 1.2.3 = 10

P(A) = n(A)/n(S) = 10/286

B -> The three students are from third year

n(B) = 4C₃ = 4

p(B) = n(B)/n(S) = 4/286

Here A and B are disjoint events since both cannot occur together.

P(AUB) = P(A) + P(B)

(10/286) + (4/286) = 14/286 = 7/143

Example 1.1.14

A bag contains 30 balls numbered from 1 to 30. One ball is drawn at random. Find the probability that the number of the ball drawn will be a multiple of (a) 5 or 7 and (b) 3 or 7.

Solution:

Given: n(S) = 30

Let A = The probability of the number being multiple of 5.

p(A) = p(5, 10, 15, 20, 25, 30) = 6/30

Let B = The probability of the number being multiple of 7.

p(B) = P(7, 14, 21, 28) = 4/30

Let C = The probability of the number being multiple of 3.

p(C) = p(3, 6, 9, 12, 15, 18, 21, 24, 27, 30) = 10/30

(a) The events A and B are mutually exclusive the probability of the number being a multiple of 5 or 7 will be

= (6/30) + (4/30) = 10/30

(b) The events C and B are not mutually exclusive.

Here p(C∩B) = p[21] = 1/30

.'. p(CUB) = p(C) + p(B) - p(C∩B)

= (10/30) + (4/30) - (1/30) = 13/30

Example 1.1.15

What is the probability of picking a card that was red or black?

Solution: Let A -> The event of picking a red card.

B -> The event of picking a black card. 26

p(A) = 26/52; p(B) = 26/52

P(AUB) = p(A) + p(B) = (26/52) + (26/52) = 1