Hello JoanF Originally Posted by

**JoanF** **In a contest, players must take 3 balls. ** **There are 3 green balls, 2 blue and one white. ** **If the third ball is white, the player pass to the second stage. ** **If the first 2 balls are the same color and the third white, the player pass automatically to the third stage.** **A player had automatically passed from first to third stage.** **What is the probability of the first 2 balls taken be blue?**
I don't have the correct solution

My answer was 2/5

Please help...

You haven't told us a vital piece of information: is the first ball returned to the bag (or whatever) after it is drawn? I will assume that it is not.

Suppose that A is the event: The first two balls are blue.

Suppose that B is the event: The player passes to the third stage.

Now the probability that the first ball drawn is blue is $\displaystyle \frac{2}{6}=\frac13$; and the probability that the second is blue (if the first is not returned to the bag) is then $\displaystyle \frac15$. So $\displaystyle p(A) = \frac13\times \frac15 = \frac{1}{15}$

Similarly the probability that the first two balls drawn are green is $\displaystyle \frac36\times\frac25=\frac15$

If the white ball has not been drawn in the first two draws, the probability that it is drawn on the third draw is $\displaystyle \frac{1}{4}$. This is also $\displaystyle p(B|A)$, the probability that the player passes to the third stage, given that the first two balls were blue.

Now event B can happen in either of two ways: where the third ball is white, with the first two balls either blue or green. So $\displaystyle p(B) = \frac{1}{15}\times \frac14 + \frac15\times\frac14 = \frac{1}{15}$

Bayes' Theorem says:$\displaystyle p(A|B) = \frac{p(B|A)p(A)}{p(B)}$

where $\displaystyle p(A|B)$ is the probability that the first two balls were blue, given that the player passed to the third stage (which is what we want). So:$\displaystyle p(A|B) = \frac{\dfrac14\times\dfrac{1}{15}}{\dfrac{1}{15}}$$\displaystyle =\frac14$

If the balls are returned to the bag each time, you can use the same method, but make adjustments to the individual probabilities.

Grandad