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Question
Ten coins are tossed. What is the probability of getting at least 8 heads?
Solution
Here, n = 10
p = `1/2`
q = `-1/2 = 1/2`
P(X ≥ 8) = P(x = 8) + P(x = 9) + P(x = 10)
= `""^10"C"_8 (1/2)^8 (1/2)^(10 - 8) + ""^10"C"_9 (1/2)^9 (1/2)^(10 - 9) + ""^10"C"_10 (1/2)^10 (1/2)^0`
= `(10!)/(8!2!) * (1/2)^8 * (1/2)^2 + (101)/(9!1!)(1/2)^9 (1/2) + (1/2)^10`
= `45*(1/2)^10 + 10*(1/2)^10 + (1/2)^10`
= `(1/2)^10 * (45 + 10 + 1)`
= `56(1/2)^10`
= `56 xx 1/1024`
= `7/128`
Hence, the required probability is `7/128`.
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