Suppose that Box-I contains 8 red,3 blue and | vedclass

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(C) Let $R_1, B_1, G_1$ be the events of choosing a red,blue,or green ball from Box-$I$ respectively. The probabilities are $P(R_1) = \frac{8}{16} = \

Vedclass · Accepted Answer

$\frac{5}{52}$

Vedclass · Answer

(C) Let $R_1, B_1, G_1$ be the events of choosing a red,blue,or green ball from Box-$I$ respectively. The probabilities are $P(R_1) = \frac{8}{16} = \frac{1}{2}$,$P(B_1) = \frac{3}{16}$,and $P(G_1) = \frac{5}{16}$.Let $A$ be the event that one of the chosen balls is white. Let $B$ be the event that at least one of the chosen balls is green.White balls are only present in Box-$IV$. Thus,$A$ can only occur if we choose a green ball from Box-$I$ and then a white ball from Box-$IV$. However,Box-$IV$ contains $10$ green,$16$ orange,and $6$ white balls (Total $32$).$P(A \cap B) = P(G_1) 	imes P(	ext{white from Box-}IV) = \frac{5}{16} 	imes \frac{6}{32} = \frac{5}{16} 	imes \frac{3}{16} = \frac{15}{256}$.Now,$P(B) = P(G_1) + P(R_1 \cap G_2) + P(B_1 \cap G_3)$,where $G_2$ is green from Box-$II$ and $G_3$ is green from Box-$III$.$P(B) = \frac{5}{16} + (\frac{8}{16} 	imes \frac{15}{48}) + (\frac{3}{16} 	imes \frac{12}{16}) = \frac{5}{16} + (\frac{1}{2} 	imes \frac{5}{16}) + (\frac{3}{16} 	imes \frac{3}{4}) = \frac{5}{16} + \frac{5}{32} + \frac{9}{64} = \frac{20+10+9}{64} = \frac{39}{64}$.$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{15/256}{39/64} = \frac{15}{256} 	imes \frac{64}{39} = \frac{15}{4 	imes 39} = \frac{5}{4 	imes 13} = \frac{5}{52}$.

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