How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%?

Explanation

Let no. of times of tossing a coin be `n`.

Here, Probability of getting a head in a chance` = p = 1/2`

Probability of getting no head in a chance` = q = 1-1/2 = 1/2`

`=1 – P(X=0) = 1 - \ ^n C_0 P^0 p^0 q^(n-0) = 1 – (1/2)^n`

From question

` 1 – (1/2)^n gt 80/100`

`Rightarrow 1 – (1/2)^n gt 8/10 Rightarrow 1 – 8/10 gt 1/2^n`

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