Coin Toss Probability Using Bayes Theorem

2.2k Views Asked by At

In a box, there are the same number of two kinds of coins: the fair coins (50% chance for head) and the biased coins (70% chance for head). One person randomly selected a coin and tosses it twice. Both tosses are tails. What is the probability that the selected coin is a biased coin?

My calculation:

$$P(\text{Fair Coin}) = 0.50,$$

$$P(\text{Biased Coin}) = 0.50,$$

$$P(\text{TT }|\text{ Fair Coin}) = 0.50,$$

$$P(\text{TT }|\text{ Biased Coin}) = 0.30.$$

$$P(\text{Biased Coin }|\text{ Tails}) = P(\text{Tails }|\text{ Biased Coin})*P(\text{Biased Coin}) / P(\text{Tails }|\text{ Biased Coin})*P(\text{Biased Coin}) + P(\text{Fair Coin})*P(\text{Tails }|\text{ Fair Coin}) $$

$$= (0.3 * 0.5) / (0.5 * 0.3) + (0.5 * 0.5) = 0.375. $$

But it does not match with the given answer, where am I going wrong?

1

There are 1 best solutions below

2
On BEST ANSWER

$F$ = fair coin, $B$ = biased coin

$P(F) = 0.5, P(B) = 0.5$

$P(TT|F) = 0.5^2 = 0.25$

$P(TT|B) = 0.3^2 = 0.09$

$P(B|TT) = \dfrac{P(TT|B)\cdot P(B)}{ P(TT|B)\cdot P(B) + P(TT|F)\cdot P(F)}$

You can compute the value

Also learn to shorten expressions by using symbols, eg $F$ and $B$ instead of writing text