Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
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? [DMRC-Customer Relationship Assistant (CRA) Exam, 2016]
(a)$\frac{1}{18}$
(b)$\frac{1}{70}$
(c)$\frac{3}{5}$
(d)$\frac{1}{2}$
Answer
Answer (as printed): B
Explanation
Number of blue balls = 3 balls Number of green balls = 2 balls Number of red balls $=5$ balls Total balls in the bag $=3+2+5=10$ Total possible outcomes = Selection of 4 balls out of 10 $$\text { balls }={ }^{10} C_{4}=\frac{10!}{4 \times(10-4)!}=\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}=210$$ Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls) $$\begin{aligned} & ={ }^{2} C_{2} \times{ }^{3} C_{2} \\ & =1 \times 3=3 \end{aligned} \begin{aligned} \therefore \text { Required probability } & =\frac{\text { Favorable out comes }}{\text { Total possible outcomes }} \\ & =\frac{3}{210}=\frac{1}{70} \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.