There are $15$ urns of $3$ different kind, filled with black and white balls, given in the table:

kind  |  amount  |  black balls  |  white balls
   I  |    2     |      10       |       15
  II  |    6     |       8       |        2
 III  |    7     |      10       |        6

One of these urns will be chosen randomly and equally distributed. From this urn, a ball is picked randomly and equally distributed. It is a black ball. What's probability that it came from an urn of kind I?

I have idea but not know if idea is good.

I first need know what is probability that black ball is from urn I.

Probability is $\frac{10}{25}$

Now need to be careful because we have in total $15$ urns and from these $15$ urns we have $2$ times the urn I.

In end we have probability to get black ball from urn I: $$\frac{2}{15} \cdot \frac{10}{25}= \frac{4}{75} \approx 0.05\bar{3} \approx 5.33\text{%}$$

Is it good or not? Pls need info for test next week.


1 Answer 1


No. You calculated the probability of getting a black ball from urn I.

What they asked for is the probability that the ball was from urn I, given that it is a black ball.

So, you calculated $P(I \cap Black)$

But they are asking for $P(I|Black)$

To calculate the latter, note that:

$$P(I \cap Black) = P(I|Black) \cdot P(Black)$$

and hence:

$$P(I|Black) = \frac{P(I \cap Black)}{P(Black)}$$

So, you already have $P(I \cap Black)$ ... now you need to get $P(Black)$

To get $P(Black)$, note that:

$$P(Black)=P(Black|I)\cdot P(I) + P(Black|II)\cdot P(II) + P(Black|III)\cdot P(III)$$

I think you can do the rest!

  • $\begingroup$ $$P(I|\text{Black}) = \frac{\frac{4}{75}}{\frac{10}{28}} \text{ ? }$$ $\endgroup$
    – eyesima
    Nov 10, 2017 at 21:53
  • $\begingroup$ @Hamudii Can you explain that? $\endgroup$
    – Bram28
    Nov 10, 2017 at 21:54
  • $\begingroup$ Ok thank you I understanded edit! $\endgroup$
    – eyesima
    Nov 10, 2017 at 21:57

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