Given two coins which have probability of getting heads p% and q% respectively, the task is to determine the probability of getting two consecutive heads after choosing random coins among the given coins.
Input: p = 33, q = 66
Input: p = 33, q = 66
Since both the coins are not identical so Bayes’s theorem will be used to get the desired probability.
As the coins are to be chosen randomly then any of them can be chosen so p and q will be included in the calculation. After applying Bayes’s theorem, the required answer will be (p * p + q * q) / (p + q) because if the first coin is chosen then the probability of getting both heads back to back is p * p and same for the second coin.
It’s an application of Bayes’ theorem.
P(B | A) = P(A |^| B) / P(A) = (1/2 * p * p + 1/2 * q * q) / (1/2 * p + 1/2 * q) = (p * p + q * q) / (p + q) where;
P(B) = probability to get heads on the second throw,
P(A) = probability to get heads on the first throw and
P(A |^| B) = probability to get heads on both the throws.
So, P(B | A) is the probability to get heads on the second throw if we are given that we got heads
on the first one.
Here A, B denotes 1st and 2nd coin.
Below is the implementation of above approach:
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