Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99 % is:

Step-by-step solution

Let coin is tossed n times. The probability of getting a head in a single throw of a die is 1 2 . By using binomial theorem, we know that the probability that an event X appear r times in n trials is P X = r = C r n p n - r q r ,   q = 1 - p ,   r = 0 ,   1 ,   2 ,   . . . ,   n . Now, P ( at least are head ) = 1 - P ( no head ) = 1 - 1 2 n Using, the given condition, we get 1 - 1 2 n ≥ 0.99 ⇒ 1 2 n ≤ 1 - 0.99 ⇒ 1 2 n ≤ 1 100 As, 2 6 = 64 and 2 7 = 128 , h