How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80% ?
How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?
-3
$\begingroup$
probability
-
5Much easier to figure out the probability of having *no* heads in $n$ throws. And, it works for women, too. – 2017-02-23
-
0https://en.wikipedia.org/wiki/Geometric_distribution – 2017-02-23
2 Answers
1
$$Pr(\text{at least one head in } n \text{ tosses}) = 1-Pr(\text{no head in } n \text{ tosses})=1-\left( \frac12 \right)^n$$
Hence we just have to solve for $$1-\left( \frac12 \right)^n > 0.8$$
$$\left( \frac12 \right)^n < 0.2$$
1
$1$ time: $50\%$
$2$ time: $(100\%-50\% \times 50\%) = 75\%$
$3$ time: $(100\%-50\% \times 50\% \times 50\%) = 87.5\%$
So answer is $3$ times