# Coin Tosses Probability

Naveen's coin box contains 8 fair standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads?

My solution: Coin tosses are independent.
P(Head) = P(Head/Fair Coin)P(Fair Coin) + P(Head/Unfair Coin)P(Unfair Coin)
= 1/2 * 8/9 + 1 * 1/9
= 4/9 + 1/9
=5/9

Am I right?

## 2 Answers

You have not made use of the fact that the coin he picked came up heads 4 times. You need to introduce an extra term in each summand of your computation for the chance that this kind of coin will come up heads four times in a row.

Guide.

Let $$H_i$$ denote the event that the $$i$$-th throw results in a head.

Let $$E$$ denote the event that a fair coin was selected.

To be found is $$P(H_5\mid \bigcap_{i=1}^4H_i)$$ which can be done on base of:

• $$P(H_5\mid \bigcap_{i=1}^4H_i)=P(\bigcap_{i=1}^5H_i)/P(\bigcap_{i=1}^4H_i)$$

• $$P(\bigcap_{i=1}^nH_i)=P(\bigcap_{i=1}^nH_i\mid E)P(E)+(\bigcap_{i=1}^nH_i\mid E^{\complement})P(E^{\complement})$$ for every $$n$$.