# 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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