Question

A box has \(6\) black, \(4\) red, \(2\) white and \(3\) blue shirts. When \(2\) shirts are picked at random, the probability that either both are white or both are blue is:

• A. \(\frac{4}{105}\)
• B. \(\frac{1}{35}\)
• C. \(\frac{1}{105}\)
• D. \(\frac{1}{15}\)

Hint:

For selection problems, first count total combinations using \(^{n}C_r\). Then count each favourable case separately and add them if they are mutually exclusive.

Answer 1

The Correct Option is A

Step 1: Find the total number of shirts.
The box contains 6 + 4 + 2 + 3 = 15 shirts. The number of ways to choose any 2 shirts is ¹⁵C₂ = (15 × 14)/2 = 105.

Step 2: Count the ways to pick two white shirts.
There are 2 white shirts, so ²C₂ = 1 way.

Step 3: Count the ways to pick two blue shirts.
There are 3 blue shirts, so ³C₂ = (3 × 2)/2 = 3 ways.

Step 4: Sum the mutually exclusive favorable cases.
Total favorable = 1 + 3 = 4.

Step 5: Determine the probability.
P = 4/105.

Step 6: Final conclusion.
The probability that both selected shirts are of the same color (both white or both blue) is 4/105.