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

Solution

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

Probability ( P ) of getting a head at the toss of a coin is $\frac{1}{2}$

$\Rightarrow p = \frac{1}{2} \Rightarrow q = \frac{1}{2} ∴ P (X - x) =^{n} C_{x} p^{n - x} q^{x} =^{n} C_{x} {(\frac{1}{2})}^{n - x} {(\frac{1}{2})}^{x} =^{n} C_{x} {(\frac{1}{2})}^{n}$

It is given that

P ( getting at least one head ) $> \frac{80}{100}$

$\Rightarrow p (X \geq 1) > 0 . 8 \Rightarrow 1 - P (X = 0) > 0 . 8 \Rightarrow 1 -^{n} C_{0} \frac{1}{2^{n}} < 0 . 8 \Rightarrow^{n} C_{0} \frac{1}{2^{n}} < 0 . 2 \Rightarrow \frac{1}{2^{n}} < 0 . 2 \Rightarrow 2^{n} > \frac{1}{0 . 2} = 5 \Rightarrow 2^{n} > 5 . . . . . . . . . . . (i)$

The minimum value of n which satisfies given inequality is 3.

Thus, the man should toss the coin 3 or more than 3 times.

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Vector Algebra

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

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