Two dice are thrown until a 6 is obtained on atleast one of them. The probability that a 6 appears at the kth throw for the first time must be

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${(\frac{25}{36})}^{k - 1} \frac{1}{6}$

${(\frac{25}{36})}^{k - 1} (\frac{11}{36})$

${(\frac{5}{6})}^{k - 1} \frac{11}{36}$

None of these

answer is B.

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Detailed Solution

p(getting atleast one 6) $= \frac{11}{36} = p$
p (not getting 6 on anydie)= $= \frac{25}{36} = q$
Req prob = $= q^{k - 1} \cdot p^{1} = {(\frac{25}{36})}^{k - 1} \cdot \frac{11}{36}$

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