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Last updated at Feb. 15, 2020 by Teachoo
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Example 35 The probability of a shooter hitting a target is 3/4 . How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?Let X : Number of times he hits the target Hitting the target is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx ๐^(๐โ๐) ๐^๐ Here, n = number of rounds fired p = Probability of hitting = 3/4 q = 1 โ p = 1 โ 3/4 = 1/4 Hence, P(X = x) = nCx (๐/๐)^๐ (๐/๐)^(๐โ๐) We need to find How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99 So, given P(X โฅ 1) > 99%, we need to find n Now, P(X โฅ 1) > 99 % 1 โ P(X = 0) > 99 % ` 1 โ nC0(3/4)^0 (1/4)^๐> 0.99 1 โ (1/4)^๐ > 0.99 1 โ 0.99 > (1/4)^๐ 0.01 > 1/4^๐ 4^๐ > 1/0.01 ๐^๐ > ๐๐๐ We know that 44 = 256 So, n โฅ 4 So, the minimum value of n is 4 So, he must fire atleast 4 times `
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