C3_W1 Quiz 1

[Musab_Bin_Gulfam]

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?

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 and Y)/P(Y)

P(X and Y) = P(X).P(Y|X) = 5000/9000 x 1500/3000 = 5/18

P(Y) = P(X).P(Y|X) + P(X’).P(Y|X’). Now how do we go from here? and is the probability calculated above correct? Please help. Thanks in advance.

[Musab_Bin_Gulfam]

So, here is the solution to the problem.

P(X|Y) = P(X∩Y)/P(Y)

P(X∩Y) = P(X) * P(Y|X)

P(X∩Y) = 5000/9000 * 1500/5000 = 15/90

P(Y) = P(X) * P(Y|X) + P(𝑋𝑐) * P(Y|𝑋𝑐)

P(Y) = 15/90 + (4000/9000 * 3000/4000)

P(Y) = 15/90 + 15/90 = 30/90

P(X|Y) = 15/90 * 90/30

P(X|Y) = 15/30 = 1/2

[omida]

I think the simplest solution, as you stated, starts with P(X|Y) = P(X∩Y)/P(Y). then,
P(Y) = P(there is a bug) is given as 3000/9000 = 1/3.
P(X∩Y) = P(version B and there’s a bug) = 1500/9000 = 1/6.
Then P(X|Y) = (1/6)/(1/3) = 0.5.

[bertie]

[quote=“Musab_Bin_Gulfam, post:2, topic:585147”]
P(X|Y) = P(X∩Y)/P(Y) P(X∩Y) = P(X) * P(Y|X) P(X∩Y) = 5000/9000 * 1500/5000 = 15/90 P(Y) = P(X) * P(Y|X) + P(𝑋𝑐) * P(Y|𝑋𝑐) P(Y) = 15/90 + (4000/9000 * 1500/4000) P(Y) = 15/90 + 15/90 = 30/90 P(X|Y) = 15/90 * 90/30 P(X|Y) = 15/30 = 1/2