A software company conducted a test on their new platform by exposing their users to two versions of the same product.
Number of users that were given version A: 4000
Number of user that were given version B: 5000
Number of users that experienced a bug: 3000
Number of users with version B that experienced a bug: 1500
What is the probability that a user tested Version B, given they experienced a bug during testing?
My attempt
I don’t get this question at all.
Let X = user tested Version B Let Y = experienced a bug during testing
$P(X|Y)=P(X \cap Y)/P(Y)$
$P(X \cap Y) = P(X) \cdot P(Y|X) = \dfrac{5000}{9000} \times \dfrac{1500}{3000} = \dfrac{5}{18}$
$P(Y) = P(X).P(Y|X) + P(X^c).P(Y|X^c)$
Now how do we go from here? and is the probability calculated above correct? Please help. Thanks in advance.
You are on the right track, except for one mistake.
$$P(Y|X) = \frac{1500}{5000} = \frac{3}{10}$$
The denominator will be $5000$ because it's the probability of a user experiencing a bug if they had version B. How many had version B? $5000$, right. How many of those experienced a bug? $1500$. See why it should be $5000$ and not $3000$?
Similarly $P(X^c) = \dfrac{4000}{9000}$ and $P(Y|X^c) = \dfrac{1500}{4000}$.
I believe you can take it from here. Just plug the values into your formula.