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

Solution:

Let us suppose that a man throws the coin n times before calling a winner. As a result, the n tosses are the Bernoulli trials.
\therefore Probability of getting head at the toss of the \operatorname{coin}=1 / 2

Let us assume, p=1 / 2 and q=1 / 2

\therefore P(X=x)={ }^{n} C_{x} p^{n-x} q^{x}

={ }^{n} C_{x}\left(\frac{1}{2}\right)^{n-x}\left(\frac{1}{2}\right)^{x}

={ }^{n} C_{x}\left(\frac{1}{2}\right)^{n}

It is given in the question that,

Probability of getting at least one head >90 / 100

\therefore P(x \geq 1)>0.9

1-P(x=0)>0.9

{ }^{1-}{ }^{n} C_{0} \cdot \frac{1}{2^{n}}>0.9

\frac{1}{2^{n}}<0.1

2^{n}>\frac{1}{0.1}

2^{n}>10

Hence, the minimum value of \mathrm{n} satisfying the given inequality =4

\therefore The man have to toss the coin 4 or more times