Question

How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?

Answer 1

Let no. of times of tossing a coin be n.

Here, Probability of getting a head in a chance = p = 1/2

Probability of getting no head in a chance = q = 1 - 1/2 = 1/2

Now, P (having at least one head) = P (X ≥ 1)

= 1 - P(X = 0) = 1 - nC0P0p0qn - 0 = 1 - 1.1.(1/2)n = 1 - (1/2)n

From question

1 - (1/2)n > 80/100

⇒ 1 - (1/2)n > 8/10 ⇒ 1 - 8/10 > 1/2n

⇒ 1/5 > 1/2n ⇒ 2n > 5 ⇒ n ≥ 3

A man must have to toss a fair coin 3 times.