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

Let the man toss the coin n times.

Let p be the probability of getting a head in a toss.

Then, q is the probability of getting a tail in a toss.

Since the coin has only two outcomes, so the probability of getting a head = 1/2 Also, p + q = 1

q = 1 – p  Let X be a random variable that represents a number of occurrence of the head in n tosses of a fair coin.

Then, the probability of getting r number of heads out of total n tosses is given by this Binomial distribution.

P (X = r) = nCrprqn-r

Substituting the value of p and q in the above equation, we get …(i)

We need to find the number of times the man must toss a fair coin so that the probability of having at least one head is more than 80%.

We can represent it as,

P (getting atleast one head) > 80%

P (X ≥ 1) > 80%

1 – P (X < 1) > 80% [ P (X ≥ 1) = 1 – P (X < 1)]

1 – P (X = 0) > 80% [ P (X < 1) = P (X = 0)]

Put r = 0 in equation (i) and then, substituting it in the above equation.         2n > 5

Now, we need to find the minimum value of n that satisfy this inequality.

Put n = 0.

20 > 5

1 > 5

But 1 5.

Put n = 1.

21 > 5

2 > 5

But 2 5.

Put n = 2.

22 > 5

4 > 5

But 4 5.

Put n = 3.

23 > 5

8 > 5

It is true.

Thus, the minimum n that satisfy this inequality is 3.

Hence, the man should toss the coin 3 or more times.

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