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A box contains $6$ red balls, $7$ green balls and $5$ blue balls. Each ball is of a different size. The probability that the red ball selected is the smallest red ball, is

1. $1/18$
2. $1/3$
3. $1/6$
4. $2/3$

Total number of balls in box $=6$ red balls $+ 7$ green balls $+ 5$ blue balls $=18$ balls

Probability of selecting red ball $=\frac{6}{18}$

The probability of selecting the smallest red ball (it is given that each ball is of different size) $=\frac{6}{18} \times \frac{1}{6}$

So probability that the red ball selected is the smallest red ball $= \dfrac{\text{Probability of red ball being selected AND the selected red ball being the smallest}}{\text{Probability of red ball being selected}}$

$\qquad \qquad = \dfrac{\frac{6}{18}\times \frac{1}{6}}{\frac{6}{18}}=\frac{1}{6}$

Option $(C)$ is correct.
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