JEE Advanced 2022 — Paper 2 — MATH-III solution

JEE Advanced2022Paper 2MATH-III

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

A
$\frac{15}{256}$
B
$\frac{3}{16}$
C
$\frac{5}{52}$
D
$\frac{1}{8}$

Solution

Probabilities of the colour of $b$ from Box-I (total 16): $P (red) = \frac{8}{16} = \frac{1}{2}$ , $P (blue) = \frac{3}{16}$ , $P (green) = \frac{5}{16}$ .

Let $A$ = 'a white ball is chosen' and $B$ = 'at least one chosen ball is green'.

White can only come from Box-IV, which requires $b$ to be green: $P (A \cap B) = P (green from I) \cdot P (white from IV) = \frac{5}{16} \cdot \frac{6}{32}$ .

$P (B)$ sums over the three cases: red from I then green from II $= \frac{1}{2} \cdot \frac{15}{48}$ ; blue from I then green from III $= \frac{3}{16} \cdot \frac{12}{16}$ ; green from I (already a green ball) $= \frac{5}{16} \cdot 1$ .

Computing: $P (B) = \frac{15}{96} + \frac{36}{256} + \frac{5}{16}$ . Common denominator $768$ gives $P (B) = \frac{120 + 108 + 240}{768} = \frac{468}{768}$ .

$P (A \cap B) = \frac{5 \cdot 6}{16 \cdot 32} = \frac{30}{512} = \frac{45}{768}$ .

Therefore $P (A ∣ B) = \frac{45/768}{468/768} = \frac{45}{468} = \frac{5}{52}$ .

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