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1. A bag contains 3 blue, 2 green and 5 red balls. If four balls are picked at random, what is the probability that two are green and two are blue?

A. $\frac{1}{18}$

B. $\frac{1}{70}$

C. $\frac{3}{5}$

D. $\frac{1}{2}$

Solution:

Number of blue balls = 3
Number of green balls = 2
Numbers of red balls = 5
Total balls in the bag = 3 + 2 + 5 = 10
Total possible outcomes = Selection of 4 balls out of 10 balls

=^{10} C_{4} = \frac{10!}{4! \times (10 - 4)!}

= \frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}

= 210
Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls)

\begin{aligned} =^{2} C_{2} \times^{3} C_{2} \\ = 1 \times 3 \\ = 3 \end{aligned}

∴ Required probability

= \frac{Favorable outcomes}{Total possible outcomes}

\begin{aligned} = \frac{3}{210} \\ = \frac{1}{70} \end{aligned}

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