Question

A drawer contains 3 white shirts, 2 blue shirts, and 5 gray shirts. A shirt is randomly
selected from the drawer and set aside. Then another shirt is randomly selected from the
drawer.
What is the probability that the first shirt is white and the second shirt is gray?

Answer 1

Answer:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = $\frac{1}{4}$

Step-by-step explanation:

Given that

3 white, 2 blue and 5 gray shirts are there.

To find:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ?

Solution:

Here, total number of shirts = 3+2+5 = 10

First of all, let us learn about the formula of an event E:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(First\ White) = \dfrac{\text{Number of white shirts}}{\text {Total number of shirts left}}$

$P(First\ White) = \dfrac{3}{10}$

Now, this shirt is set aside.

So, total number of shirts left are 9 now.

$P(First\ White\ and\ second\ gray) = P(First White) \times P(Second\ Gray)\\\Rightarrow P(First\ White\ and\ second\ gray) = P(First White) \times \dfrac{\text{Number of gray shirts}}{\text{Total number of shirts left}}\\\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{3}{10} \times \dfrac{5}{9}\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow P(First\ White\ and\ second\ gray) = \bold{\dfrac{1}{4} }$

So, the answer is:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = $\frac{1}{4}$