Question

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

Answer 1

Let the required coin toses be n

Pr(at least one head in n tosses)>0.8 

                  or

Pr(0 heads in n tosses)=<0.2

Probability of 0 heads in n tosses is (1/2)ⁿ

(1/2)ⁿ =< 0.2 take log on both sides

nlog(0.5)=<log(0.2)

n*(-0.3010)=<(-0.6989) or n=>2.3219

∴ approximately it's 3

Answer 2

Let x be the number of heads appearing.

Let n be the number of coins tosses.

In any fair coin toss, P(getting a head) = P(getting a tail) 

                                  p = q = 1/2.


We need to find n such that the probability of getting at least one head is more than 80%.

P(x>1)>80%

1 - P(x = 0) > 80%

1 - nc0(1/2)^n > 80%

$1 - \frac{1}{2^n} \ \textgreater \ 80%$%

$1 - \frac{1}{2^n} \ \textgreater \ \frac{80}{100}$

$1 - \frac{1}{2^n} \ \textgreater \ \frac{8}{10}$

$1 - \frac{8}{10} \ \textgreater \ \frac{1}{ 2^{n} }$

$\frac{10 - 8}{10} \ \textgreater \ \frac{1}{ 2^{n} }$

$\frac{2}{10} \ \textgreater \ \frac{1}{ 2^{n} }$

$\frac{1}{5} \ \textgreater \ \frac{1}{2^{n} }$

$2^{n} \ \textgreater \ 5$

2^3 = 8

8 > 5.


The value of n = 3.

Hence, the man tosses at least 3 times.


Hope this helps!