How many times must a man toss a fair com so that the probability of having at least one head is more than 90%?

Answer

N/a

Explanation

Let a man tosses an unbiased coin `n` times and the probability of getting a head in one toss
`=1/2`
`p=1/2` and `q=1-1/2=1/2`
`:.P(X=r)=.^(n)C_(r)p^(r)q^(n-r)`
`=.^(n)C_(r)(1/2)^(r)(1/2)^(n-r)=.^(n)C_(2)(1/2)^(n)`
Given that `P` (getting at least one head) `gt90%`
`implies1-P(0)gt90/100implies1-.^(n)C_(0)p^(0)q^(n)gt9/10`
`implies1-^(n)C_(0)(1/2)^(0)(1/2)^(n)gt9/10`
`1-9/10gt1/(2^(n))`
`implies2^(n)gt10impliesngt4`
Therefore a man must toss an unbiased coin at least 4 times.