A fair die is tossed eight times. Probability that on the eighth throw a third six is observed is
- $\displaystyle^{ 8 }{ { C }_{ 3 } }\frac { { 5 }^{ 8 } }{ { 6 }^{ 8 } } $
- $\displaystyle^{ 7 }{ { C }_{ 2 } }\frac { { 5 }^{ 5 } }{ { 6 }^{ 8 } } $
- $\displaystyle^{ 7 }{ { C }_{ 2 } }\frac { { 5 }^{ 5 } }{ { 6 }^{ 7 } } $
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None of these
For the third six to occur on the eighth throw, exactly two sixes must occur in the first seven throws, and the eighth throw must be a six. The probability of getting two sixes in seven throws is 7C2 * (1/6)^2 * (5/6)^5. Multiplying this by the probability of a six on the eighth throw (1/6) gives 7C2 * (5^5 / 6^8).
To get a third six exactly on the eighth throw, there must be exactly two sixes in the first seven throws, and the eighth throw must be a six. The probability of exactly two sixes in seven throws follows the binomial distribution, calculated as 7C2 times (1/6)^2 times (5/6)^5. Multiplying this by the 1/6 probability of rolling a six on the eighth throw gives 7C2 times (1/6)^2 times (5/6)^5 times 1/6. Combining the terms results in 7C2 multiplied by 5^5 divided by 6^8.