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If the ball is either blue or white, what is the probability that it was drawn from urn Y?
Assume that you randomly pick one of the urns: X, Y, or Z. You then randomly draw one ball out of the selected urn and note its color. The urns contain the following colored balls:

Urn X: 2 red, 1 white, and 2 blue.
Urn Y: 1 red, 2 white, and 3 blue.
Urn Z: 1 red, 2 white, and 1 blue.

If the ball is either blue or white, what is the probability that it was drawn from urn Y?

For this problem I did this and I can't figure out why it is wrong:

Pr(drawn from Y urn/is blue) + Pr(drawn from Y urn/ is
white)

(10/23)+(10/31)=540/713

The probability of an event E occurring is 0.8. What are the odds in favor of E occurring?

To find the probability that the ball was drawn from urn Y given that the ball is either blue or white, you need to use Bayes' theorem.

Let's break down the problem step by step:

1. Determine the probability of drawing a blue or white ball:
- In urn X: 2 blue + 1 white = 3 balls
- In urn Y: 3 blue + 2 white = 5 balls
- In urn Z: 1 blue + 2 white = 3 balls
Total number of blue or white balls = 3 + 5 + 3 = 11

So, the probability of drawing a blue or white ball is 11 out of the total number of balls in all three urns, which is 11/23.

2. Determine the probability of drawing a blue or white ball from urn Y:
- In urn Y: 3 blue + 2 white = 5 balls
So, the probability of drawing a blue or white ball from urn Y is 5 out of the total number of balls in urn Y, which is 5/6.

3. Apply Bayes' theorem:
Bayes' theorem states that the probability of an event A happening given that event B has already occurred can be calculated as:
P(A|B) = (P(B|A) * P(A)) / P(B)

In this case, event A is drawing the ball from urn Y, and event B is drawing a blue or white ball.

P(A) is the prior probability of drawing from urn Y, which is 1/3 (since there are 3 urns and the selection was random).
P(B|A) is the probability of drawing a blue or white ball from urn Y, which is 5/6 (as calculated in step 2).
P(B) is the probability of drawing a blue or white ball, which is 11/23 (as calculated in step 1).

Plugging these values into the equation, you get:
P(Y|blue or white) = (5/6 * 1/3) / (11/23)
= (5/18) / (11/23)
= (5/18) * (23/11)
= 115/198

Therefore, the probability that the ball was drawn from urn Y given that it is either blue or white is 115/198.