TITLE: Find probability of biased Coins
QUESTION [6 upvotes]: There are two biased coins A and B. We have been given the following:  
P(H|Coin A) = 0.9
P(T|Coin A) = 0.1
P(H|Coin B) = 0.1
P(T|Coin B) = 0.9  
We are also given that 
Probability of tossing Coin A or Coin B randomly is 0.5
How could we find the probability of "11th toss = Head", given that we have 10 heads in a row ?

REPLY [3 votes]: Given we have coin A, the probability of tossing 10 Heads in a row is $0.9^{10}$;
given we have coin B, the probability of tossing 10 Heads in a row is $0.1^{10}$. Using Bayes rule we have
$$P(A|10H)=\frac{P(A)P(10H|A)}{P(10H)}=\frac{0.5\cdot 0.9^{10}}{0.5\cdot 0.1^{10}+0.5\cdot 0.9^{10}}=\frac{0.9^{10}}{0.1^{10}+0.9^{10}}$$
$$P(B|10H)=\frac{P(B)P(10H|B)}{P(10H)}=\frac{0.5\cdot 0.1^{10}}{0.5\cdot 0.1^{10}+0.5\cdot 0.9^{10}}=\frac{0.1^{10}}{0.1^{10}+0.9^{10}}$$
The probability $P$ that the 11th toss is Heads given 10 previous tosses were Heads is
$$P=P(A|10H)\cdot 0.9 + P(B|10H)\cdot 0.1$$