Selecting a Sweater At a men’s clothing store, 12 men purchased blue golf sweaters, 8 purchased green sweaters. 4 purchased gray sweaters, and 7 bought black sweaters. If a customer is selected at random, find the probability that he purchased

a. A blue sweater

b. A green or gray sweater

c. A green or black or blue sweater

d. A sweater that was not black

Step 1 of 3:

In a men’s clothing store, 12 men purchased blue golf sweaters, 8 purchased green sweaters.

Four of them purchased gray sweaters, and 7 bought black sweaters.

A customer is randomly selected.

We have to find the following probabilities.

A blue sweater.A green or gray sweater.A green or black or blue sweater.A sweater that was not black.Step 2 of 3:

Let us define the events,

A: {Event that selecting a blue sweater.}

B: {Event that selecting a green sweater.}

C: {Event that selecting a gray sweater.}

D: {Event that selecting a black sweater.}

Total number of men = 12+ 8+ 4+ 7

= 31

It is given that, in this 12 men purchased blue golf sweaters, 8 purchased green sweaters.

Four of them purchased gray sweaters, and 7 bought black sweaters.

Here all events are mutually exclusive (it is not possible to occur any two events simultaneously).

P (AB) = 0

P (BC) = 0

P (AD) = 0

P (BD) = 0

P (ABD) = 0

By the classical definition of probability,

Probability = .

(a)

The probability of selecting a blue sweater.

P (A) =

= 0.3870

Therefore, the probability of selecting a blue sweater is 0.3870.

(b)

The probability of selecting a green or gray sweater.

Required probability = P (BC)

= P (B) + P (C)- P (BC) [Addition property]

P (Selecting a green sweater) = P (B)

=

P (Selecting a gray sweater) = P (C)

=

P (Selecting a green and a gray sweater) = P (BC)

= 0 [ Mutually exclusive events]

P (Selecting a green or a gray sweater) = P (BC)

= + -

=

= 0.3870

Therefore, the probability of selecting a green or a gray sweater is 0.3870.