A bag contains 3 red, 2 blue, 5 green and 4 yellow balls. If two balls are picked at random, what is the probability that either both are red or both are blue?

4/91

The correct answer is Option (3) → 4/91

Step 1: Total number of balls

$3 \, (\text{red}) + 2 \, (\text{blue}) + 5 \, (\text{green}) + 4 \, (\text{yellow}) = 14 \text{ balls}$

Total ways to pick 2 balls:

$\begin{pmatrix}14\\2\end{pmatrix} = \frac{14 \cdot 13}{2} = 91$

Step 2: Favorable outcomes

Case 1: Both red

$\begin{pmatrix}3\\2\end{pmatrix} = 3$

Case 2: Both blue

$\begin{pmatrix}2\\2\end{pmatrix} = 1$

Total favorable outcomes = 3 + 1 = 4

Step 3: Probability

$P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{4}{91}$