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A box ' $A$ ' contanis 2 white, 3 red and 2 black balls. Another box ' $B^{\prime}$ contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box ' $B^{\prime}$ is
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Verified Answer
The correct answer is:
$\frac{7}{16}$
$\frac{7}{16}$
Probability of drawing a white ball and then a red ball from bag $B$ is given by
$$
\frac{{ }^4 C_1 \times{ }^2 C_1}{{ }^9 C_2}=\frac{2}{9}
$$
Probability of drawing a white ball and then a red ball from bag $A$ is given by
$$
\frac{{ }^2 C_1 \times{ }^3 C_1}{{ }^7 C_2}=\frac{2}{7}
$$
Hence, the probability of drawing a white ball and then a red ball from bag $B$
$$
=\frac{\frac{2}{9}}{\frac{2}{7}+\frac{2}{9}}=\frac{2 \times 7}{18+14}=\frac{7}{16}
$$
$$
\frac{{ }^4 C_1 \times{ }^2 C_1}{{ }^9 C_2}=\frac{2}{9}
$$
Probability of drawing a white ball and then a red ball from bag $A$ is given by
$$
\frac{{ }^2 C_1 \times{ }^3 C_1}{{ }^7 C_2}=\frac{2}{7}
$$
Hence, the probability of drawing a white ball and then a red ball from bag $B$
$$
=\frac{\frac{2}{9}}{\frac{2}{7}+\frac{2}{9}}=\frac{2 \times 7}{18+14}=\frac{7}{16}
$$
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