Question

In a set of 10 coins, 2 coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides.
or
How many times must a fair coin be tossed so that probability of getting at least one head is more than 80%?

Answer 1

This is based on Bayes Theorem
Let $E_1$, be the event that the coins with both heads are selected.
$E_2$ be the event that normal coins are selected A be the event that only heads are the result
probability of $E_1=\frac{2}{10}$ (2 coins out of 10)
Probability of $E_2=\frac{2}{10}$ (8 coins out of 10)
we need to find out probability of $E_1$ with geeting heads is all tosses = $P\left(\frac{E_1}{A}\right)$
Now $P\left(\frac{A}{E_1}\right)$ =prob of getting all heads with $E_1$ selected =1 (obsious since coin has heads as both sides)
$P\left(\frac{A}{E_2}\right)=(\frac{1}{2}{)}^5=\frac{1}{25}(\frac{1}{2}$ is prob of getting heads is every toss & them are 5 tuosses)
From bases theorem
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)=P\left(E_2\right)P\left(\frac{A}{E_2}\right)}=\frac{\frac{2}{10}\times 1}{\frac{2}{10}\times 1+\frac{2}{10}\times \frac{1}{2}s}=\frac{\frac{2}{10}}{\frac{64+8}{320}}=\frac{\frac{2}{10}}{\frac{36}{32}}=\frac{32}{36}=\frac{8}{9}$