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Question

# A die is thrown three times. Find P (A/B) and P (B/A), if A = 4 appears on the third toss, B = 6 and 5 appear respectively on first two tosses.

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Solution

## Consider the given events. A = Getting 4 on third throw B = Getting 6 on first throw and and 5 on second throw Clearly, A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4), . . . . . . (6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)} B = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)} $\mathrm{Now},\phantom{\rule{0ex}{0ex}}A\cap B=\left\{\left(6,5,4\right)\right\}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Required}\mathrm{probability}=P\left(A/B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Required}\mathrm{probability}=P\left(B/A\right)=\frac{n\left(A\cap B\right)}{n\left(A\right)}=\frac{1}{36}$

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