A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is

Solution :

Probability of getting success, p = \(1\over 6\)

and probability of failure, q = \(5\over 6\)

Now, we must get 2 sixes in seven throws, so probability is \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)

and the probability that 8th throw is \(1\over 6\).

\(\therefore\)   Required Probability = \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)\(1\over 6\)

= \(^7C_2\times 5^5\over {6^8}\)


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