JEE Advanced 2022 Paper 2 · Q17 · Conditional Probability

Question

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

SolveKar AI · Accepted Answer

Box-I totals: 8+3+5=16. $P(b=R)=1/2$, $P(b=B)=3/16$, $P(b=G)=5/16$. White appears only in Box-IV (6/32 = 3/16); reached only when $b$ is green. So $P(	ext{white and at least one green}) = P(b=G)\cdot P(	ext{white from IV}) = \dfrac{5}{16}\cdot\dfrac{6}{32} = \dfrac{30}{512}=\dfrac{15}{256}$. Compute $P(	ext{at least one green})$: green from box-I directly (with the second ball being anything) contributes $\dfrac{5}{16}\cdot 1=\dfrac{5}{16}$; if $b$ red, second from Box-II green: $\dfrac{1}{2}\cdot\dfrac{15}{48}=\dfrac{1}{2}\cdot\dfrac{5}{16}=\dfrac{5}{32}$; if $b$ blue, second from Box-III green: $\dfrac{3}{16}\cdot\dfrac{12}{16}=\dfrac{36}{256}=\dfrac{9}{64}$; if $b$ green, second from Box-IV green: included via $\dfrac{5}{16}\cdot\dfrac{10}{32}=\dfrac{50}{512}$ — but we already counted the first ball being green. Total = $\dfrac{5}{16}+\dfrac{5}{32}+\dfrac{9}{64} = \dfrac{20+10+9}{64}=\dfrac{39}{64}$. So conditional probability $= \dfrac{15/256}{39/64} = \dfrac{15}{256}\cdot\dfrac{64}{39}=\dfrac{15}{4\cdot 39}=\dfrac{15}{156}=\dfrac{5}{52}$. Option (C).

JEE Advanced 2022 Paper 2 · Q17 · Conditional Probability

Solution