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Question

A box contains 2 black, 6 green and 4 yellow balls. If 2 balls are picked up at random, the probability that both are green is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\frac{5}{22}$

Analyzing the Probability Problem

The question asks for the probability of selecting two green balls when drawing two balls randomly from a box containing balls of different colors. We need to calculate the number of ways to choose 2 green balls versus the total number of ways to choose any 2 balls.

Calculating Probability of Green Balls

Total Number of Balls

First, find the total number of balls in the box:

Total balls = (Number of black balls) + (Number of green balls) + (Number of yellow balls)

Total balls = 2 + 6 + 4 = 12

Total Possible Outcomes

Calculate the total number of ways to choose 2 balls from the 12 available balls. This uses combinations, denoted as $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:

Total combinations = $\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12!}{2!10!} = \frac{12 \times 11}{2 \times 1} = 66$

Favorable Outcomes (Both Green)

Calculate the number of ways to choose 2 green balls from the 6 green balls available:

Favorable combinations = $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$

Probability Calculation

The probability of an event is the ratio of favorable outcomes to total possible outcomes:

Probability (both green) = $\frac{\text{Favorable combinations}}{\text{Total combinations}}$

Probability (both green) = $\frac{15}{66}$

Simplifying the Probability

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:

Probability (both green) = $\frac{15 \div 3}{66 \div 3} = \frac{5}{22}$

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