InterviewSolution
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InterviewSolution
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The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is :A. 7B. 9C. 8D. 5 |
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Answer» Correct Answer - C Let the coin be tossed n times and let X-denote the number of heads in n tosses of the coin. Then, `therefore P(X=r)= .^(n)C_(r )((1)/(2))^(r )((1)/(2))^(n-r) = .^(n)C_(r )((1)/(2))^(n), r=0,1,2,..,n` It is given that `P(X ge 2) ge 0.96` `implies underset(r=2)overset(n)(sum)P(X=r) ge 0.96` `implies underset(r=2)overset(n)(sum) .^(n)C_(r )((1)/(2))^(n) ge 0.96` `implies (1)/(2^(n))( underset(r=2)overset(n)(sum) .^(n)C_(r )) ge 0.96` `implies (1)/(2^(n)) (2^(n) - .^(n)C_(0) - .^(n)C_(1)) ge 0.96` `implies 1-(n+1)/(2^(n)) ge 0.96` `implies (n+1) le (0.04) 2^(n)` `implies n=8, 9, 10,`.. |
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