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An urn contains four blue balls and three white balls. A second urn contains five blue and four white balls. Pass up a ball from the first to the second urn and then extracted a ball second urn. How likely to be white?

If the result is blue ball, what is the probability that the first recession resulted white ball?

I tried:

  • if the ball is white -probability of being White in the second is $\frac12$ we can have this 3 times

  • else, the probability of being white is $\frac25$ we can have this $4$ times

So probability of being white is $\frac12\cdot \frac37=\frac{3}{14}$?

If we get one blue and the firt one was white... we could get $\frac{3}{14}\cdot \frac12 = \frac3{28}$.

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  • $\begingroup$ The title of this question sounds like a good night in. $\endgroup$ – Alec Teal Sep 3 '15 at 11:28
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You made quite a couple of mistakes, but the actually, you only made one crutial mistake: You did not go about your problem systematically. You throw a lot of numbers around, but I advise you that you rather spend more time thinking about which number goes where. It will help you solve more difficult tasks further on.

OK, let's start from the beginning. We have a two-step process. In the first step, one of the two events occurs: $H_1$ is the event "a white ball is taken from the first urn", and $H_2$ is the event "a blue ball is taken from the first urn."

It is fairly obvious that $H_1$ and $H_2$ are mutually exclusive and cover the whole space of possible events, i.e. $\{H_1,H_2\}$ is a partition of the space of possible events in the first step.

Then, in the second step, we want to know the probability of the event $X$ which is "a white ball is taken from the second urn" .

You can write the probability of $P(X)$ as

$$P(X) = P(X|H_1)\cdot P(H_1) + P(X|H_2)\cdot P(H_2)$$

Now, you simply need to calculate the probabilities, which you already did, and plug them into the equation on the correct places (which you did not).

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  • $\begingroup$ P(H1)= 3/7 and P(H2)=4/7 OK. P(X|H1)=3/7*(1/2+2/5) and P(X|H2)=4/7*(1/2+3/5)? $\endgroup$ – Jean Dennis Sep 3 '15 at 11:45
  • $\begingroup$ @JeanDennis How did you get $P(X|H_1)$? Remember, $P(X|H_1)$ is the probability that you will draw a white ball from the second urn after you already draw a white ball from the first urn. $\endgroup$ – 5xum Sep 3 '15 at 11:49
  • $\begingroup$ P(X|H1)= If the first was not white, 4/7 multiplied by 4/10 (the probability of being white). If it was, 3/7 multiplied by 5/10 (the probability of being white). So, P(X|H1)= 4/7 *4/10 +3/7*5/10 ? $\endgroup$ – Jean Dennis Sep 3 '15 at 12:01
  • $\begingroup$ @JeanDennis What you wrote is the formula for $P(X)$, not $P(X|H_1)$, but yes, you are correct! $\endgroup$ – 5xum Sep 3 '15 at 12:03
  • $\begingroup$ the probability of being blue is 39/70... because it is 1-(31/70)=39/70. If the first is white the probability is 3/7. So If the result is blue ball, the probability that the first recession resulted white ball is 39/70*3/7? $\endgroup$ – Jean Dennis Sep 3 '15 at 12:21

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