How many times must a man toss a fair coin so that the probability of having at least one head - Maths - Probability

Question

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

Shridhar Kaushik · Accepted Answer

Let p denote the probability that the man will get the head.

$Then, p = \frac{1}{2} and q = 1 - \frac{1}{2} = \frac{1}{2}$

Suppose the man tosses the coin n times. Let X denotes the number of times he get head in n-trials.

$∴ P (X = r) = n_{c_{r}} p^{r} q^{n - r} = n_{c_{r}} (\frac{1}{2})^{r} (\frac{1}{2})^{n - r} = n_{c_{r}} (\frac{1}{2})^{n}, r = 0, 1, ..., n$

Now,

$P (X \geq 1) > \frac{80}{100} [Given] \Rightarrow P (X = 1) + P (X = 2) + ... + P (X = n) > \frac{8}{10} \Rightarrow [P (X = 0) + P (X = 1) + ... + P (X = n)] - P (X = 0) > \frac{8}{10} \Rightarrow 1 - P (X = 0) > \frac{8}{10} [∵ Sum of probability is 1] \Rightarrow P (X = 0) < 1 - \frac{8}{10} \Rightarrow P (X = 0) < \frac{2}{10} \Rightarrow n_{c_{0}} (\frac{1}{2})^{n} < \frac{1}{5} \Rightarrow 1 \cdot (\frac{1}{2})^{n} < \frac{1}{5}$

Now, $(\frac{1}{2})^{n} < \frac{1}{5}$ is true only when the value of n is equal to or greater than 3

$∴ n_{c_{0}} (\frac{1}{2})^{n} < \frac{1}{5} \Rightarrow n = 3, 4, 5, ...$

Hence, the man must toss at least 3 times

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