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

Explanation

Let the man toss the coin `n` times. The `n` tosses are `n` Bernoulli trials.

Probability `(p)` of getting a head at the toss of a coin is `frac{1}{2}`

`therefore p=frac{1}{2} Rightarrow q=frac{1}{2} `

`therefore {P}({X}=x)=^{n} {C}_{x} p^{n-x} q^{x}=^{n} {C}_{x}(frac{1}{2})^{n-x}(frac{1}{2})^{x}=^{n} {C}_{x}(frac{1}{2})^{n}`

It is given that,

`P(text { getting at least one head })>frac{90}{100} `

`P(x geq 1)>0.9 `

`Rightarrow 1-P(x=0)>0.9`
`1-^nC_0 cdot frac{1}{2^{n}}>0.9`

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