Box-I contains 3 cards bearing numbers 1, 2, | vedclass

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(B) Let $O_i$ and $E_i$ denote the events of drawing an odd and an even number from Box-$i$ respectively.For Box-$I$ $(1, 2, 3)$: $P(O_1) = \frac{2}{3

Vedclass · Accepted Answer

$\frac{53}{105}$

Vedclass · Answer

(B) Let $O_i$ and $E_i$ denote the events of drawing an odd and an even number from Box-$i$ respectively.For Box-$I$ $(1, 2, 3)$: $P(O_1) = \frac{2}{3}$,$P(E_1) = \frac{1}{3}$.For Box-$II$ $(1, 2, 3, 4, 5)$: $P(O_2) = \frac{3}{5}$,$P(E_2) = \frac{2}{5}$.For Box-$III$ $(1, 2, 3, 4, 5, 6, 7)$: $P(O_3) = \frac{4}{7}$,$P(E_3) = \frac{3}{7}$.The sum $x_1+x_2+x_3$ is odd if we have either three odd numbers or one odd and two even numbers.Case $1$: All three are odd: $P(O_1 \cap O_2 \cap O_3) = \frac{2}{3} 	imes \frac{3}{5} 	imes \frac{4}{7} = \frac{24}{105}$.Case $2$: One odd and two even:- $O_1, E_2, E_3$: $\frac{2}{3} 	imes \frac{2}{5} 	imes \frac{3}{7} = \frac{12}{105}$.- $E_1, O_2, E_3$: $\frac{1}{3} 	imes \frac{3}{5} 	imes \frac{3}{7} = \frac{9}{105}$.- $E_1, E_2, O_3$: $\frac{1}{3} 	imes \frac{2}{5} 	imes \frac{4}{7} = \frac{8}{105}$.Total Probability $= \frac{24+12+9+8}{105} = \frac{53}{105}$.

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