Probability of picking a biased coin Suppose you have a bag of 100 coins of which 1 is biased with both sides as Heads. You pick a coin from the bag and toss it three times. The result of all three tosses is Heads. What is the probability that the selected coin is biased?
My answer:-
P(selecting a biased coin) = 1/100
P(getting a head thrice with the biased coin) = 1

P(selecting an unbiased coin) = 99/100
P(getting a head thrice with the unbiased coin) = 1/8

P(selecting a biased coin|coin toss resulted in 3 heads) = 
          P(selecting a biased coin and getting heads thrice)/
          [P(selecting a biased coin and getting heads thrice) +
           P(selecting an unbiased coin and getting heads thrice)] 

= (1/100)/[(1/100) + (99/800)]
= (1/100)/(107/800)
= 8/107
= 0.0747

Is this correct? Thanks.
 A: Your answer is right. The solution can be derived using Bayes' Theorem:
$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$
You want to know the probability of $P(\text{biased coin}|\text{three heads})$.
What do we know?
There are $100$ coins. $99$ are fair, $1$ is biased with both sides as heads.
With a fair coin, the probability of three heads is $0.5^3 = 1/8$.
The probability of picking the biased coin: $P(\text{biased coin}) = 1/100$.
The probability of all three tosses is heads: $P(\text{three heads}) = \frac{1 \times 1+ 99 \times \frac{1}{8}}{100}$.
The probability of three heads given the biased coin is trivial: $P(\text{three heads}|\text{biased coin}) = 1$.
If we use Bayes' Theorem from above, we can calculate 
$$P(\text{biased coin}|\text{three heads}) = \frac{1 \times 1/100}{\frac{1 + 99 \times \frac{1}{8}}{100}} = \frac{1}{1 + 99 \times \frac{1}{8}} = \frac{8}{107} \approx 0.07476636$$ 
A: Here is a write that describes something very similar to that. The Bayes approach is the right way to proceed.
A: In general,
P(biased coin|k heads) = (2^k)/[(2^k) + 99]
Where k is no of consective heads
so if the trick coin was tossed 3 times
(2^3)/[(2^3) + 99] = 8/(8+99) = 8/107 = 0.07
