JEE Advanced 2022 · Paper 2 · Mathematics Q17

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MathematicsSingle correct+3 / -1Question 17 of 54

Suppose that

Box-I contains 8 red, 3 blue and 5 green balls,
Box-II contains 24 red, 9 blue and 15 green balls,
Box-III contains 1 blue, 12 green and 3 yellow balls,
Box-IV contains 10 green, 16 orange and 6 white balls.

A ball is chosen randomly from Box-I; call this ball $b$ . If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

\dfrac{15}{256}

\dfrac{3}{16}

\dfrac{5}{52}

\dfrac18

Answer and worked solution

Official answer: C

At least one green.

$b$ green: $\frac5{16}$ .
$b$ red, then green from Box-II: $\frac8{16}\cdot\frac{15}{48}=\frac{15}{96}$ .
$b$ blue, then green from Box-III: $\frac3{16}\cdot\frac{12}{16}=\frac9{64}$ .

Total: $\frac{60+30+27}{192}=\frac{117}{192}$ .

White and at least one green. White exists only in Box-IV, which is used only when $b$ is green:
$\frac5{16}\cdot\frac6{32}=\frac{15}{256}.$

Answer.
$\frac{15/256}{117/192}=\frac{15\cdot192}{256\cdot117}=\frac5{52}.$
(C).

Topics:Conditional probability Bayes' theorem

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JEE Advanced 2022 · Paper 2, Mathematics question 17