Suppose you have 3 jars with the following contents...

Question

Jar 1 has 1white ball and 22 black balls.

Jar 2 has 2 white balls and 3 black balls.

Jar 3 has 2 white balls and 1 black ball.

One jar is to be ​selected, and then 1 ball is to be drawn from the selected jar. The probabilities of selecting the​ first, second, and third jars are 1/3​, 1/2​, and 1/6 respectively. Find the probability the ball was drawn from Jar

2​, given that the ball is white.

Answer 1

Let's call the event that the ball was drawn from jar 2 "J2" and the event it's white "W".

In these terms, the problem is asking for P(J2 | W) .

For conditional probability, we use Bayes' Theorem.

P(J2 | W) = P(J2) * P(W | J2) / P(W)

We can plug in values from the given information easily for the numerator. P(J2) is 1/2. P(W | J2) is 2/5 (2 of 5 balls are white in J2).

For the denominator, P(W), we must get the overall chance we got a white ball. This is equal to P(J1)*P(W | J1) + P(J2)*P(W | J2) + P(J3)*P(W | J3), or more simply, the sum of the chance we get each jar times the chance we get a white ball from that jar.

Finally, just take the numerator over the denominator for our answer.

Not sure if the 22 is a typo or not (seems odd), so I'll let you do the actual computation, and please let me know if this answer uses too much shorthand or skips over anything you don't understand!