Quantitative Reasoning

Probability

Medium

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?

1/7

3/91

4/91

5/91

✅ Correct Option: 3

The bag contains:

Red balls: 3
Blue balls: 2
Green balls: 5
Yellow balls: 4

Total balls:

$3 + 2 + 5 + 4 = 14$

The total number of ways to pick 2 balls from 14 balls:

$C(14, 2) = \dfrac{14 \times 13}{2}$

$= \dfrac{182}{2}$

$= 91$

For both balls to be red, the number of ways to pick 2 red balls from 3 red balls:

$C(3, 2) = \dfrac{3 \times 2}{2} = 3$

For both balls to be blue, the number of ways to pick 2 blue balls from 2 blue balls:

$C(2, 2) = 1$

Total favorable outcomes:

$3 + 1 = 4$

The probability that either both are red or both are blue:

$P = \dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

$= \dfrac{4}{91}$

Therefore, the probability is $\dfrac{4}{91}$ .

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Solution

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