The minimum number of times one has to toss fair coins so that the probability of observing at least one head is at least 90%

Correct option is D. 4
Let the man toss the coin n times. Then tosses are Bernoulli trials.
Probability (p) of getting a head at the toss of a coin is 1/2
p = 1/2, q = 1/2
∴P(X = x) = nCxp^(n-x)q^x = nCx(1/2)^(n-x)(1/2)^x = nCx(1/2)^n
It is given that,
P(getting at least one head) > 90/100
P(X ≥ 1) > 0.9
⇒ 1 - P(X = 0) > 0.9
1 - nC0 * (1/2)^n > 0.9
nC0 * (1/2)^n < 0.1
(1/2)^n < 0.1
2^n > 10
(1)
The minimum value of n that satisfies the given inequality is 4.
Thus, the man should toss the coin 4 or more than 4 times.