We have two urns, one with 5 red and 5 blue colored balls, the second with 2 red and 6 blue colored balls.

Choosing randomly one of the urns and taking out one ball, what is the probability to take out a red ball ?

What I did is: $\frac 1 2 (\frac 1 4 + \frac 2 4)=\frac 3 8$ but it's apparently wrong. What is the right approach ?


Let $A$ denote the event of taking out a red ball and $B$ denote the event of choosing the first urn.

By the law of total probability, $$ \Pr(A)=\Pr(A\mid B)\Pr(B)+\Pr(A\mid B^c)\Pr(B^c) $$ and $$ \Pr(A)=\frac12\Bigl(\frac12+\frac14\Bigr)=\frac38. $$

  • $\begingroup$ So I was right and the book was wrong. Thanks. $\endgroup$ – GinKin Mar 25 '14 at 13:57
  • $\begingroup$ @GinKin You're welcome! $\endgroup$ – Cm7F7Bb Mar 25 '14 at 14:04

I disagree.

You can't simply average the two individual probabilities. Since the two urns are picked at random the probability is exactly the same as if all the balls were in one urn, i.e. 7/18.

The difference arises because there are not the same number of balls in each, so the hard way is is to take 10/18*1/2 + 8/18 *1/4 but the answer is the same.

  • $\begingroup$ It's not an average, it's the 50% chance to choose one of the two urns. There's a plus between the two urns because there's $OR$ between them. $\endgroup$ – GinKin Mar 25 '14 at 14:20
  • $\begingroup$ No you are right. I am wrong. $\endgroup$ – Dave Mear Mar 25 '14 at 14:23
  • $\begingroup$ OK, it confused me as well. $\endgroup$ – GinKin Mar 25 '14 at 14:24
  • 1
    $\begingroup$ @DaveMear It looks like just a simple average but it actually is $\frac12\cdot p+\frac14\cdot(1-p)$ and "choosing randomly" means that $p=\frac12$. $\endgroup$ – Cm7F7Bb Mar 25 '14 at 14:25

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