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.