Question:
How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?
Answer:
Let the man toss the coin n times.
Probability of getting a head in one toss P =1/2
Probability of getting a tail in one toss q =1 - 1/2 = 1/2
Let X represents a random variable having one head.
Now n tosses are n Bernoulli trials.
Now P(X = x) = nCx * pn-x * qx = nCx * (1/2)n-x * (1/2)x = nCx *(1/2)n
From question, it is given that
Probability of getting at least one head > 90/100 = 9/10
=> P(x > = 1) > 0.9
=> 1 - P(x = 0) > 0.9
=> 1 - 0.9 > P(X = 0)
=> P(X = 0) < 0.1
=> nC0 *(1/2)n < 0.1
=> (1/2)n < 0.1
=> 2n > 1/0.1
=> 2n > 10
The minimum value of n satisfies this equation is n = 4.
So the man should toss the coin at least 4 times.
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