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Question
Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.
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Solution
Here success is a score which is a multiple of 3
i.e., 3 or 6.
Therefore, P(3 or 6) = `2/6 = 1/3`
The probability of r successes in 10 throws is given by
P(r) = `""^10"C"_"r" (1/3)^"r" (2/3)^(10 - "r")`
Now P(At least 8 successes) = P(8) + P(9) + P(10)
= `""^10"C"_8 (1/3)^8 (2/3)^2 + ""^10"C"_9 (1/3)^9 (2/3)^1 + ""^10"C"_10 (1/3)^10`
= `1/3^10 [45 xx 4 + 10 xx 2 + 1]`
= `201/3^10`.
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