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An urn contains $10$ black and $5$ white balls. Two black balls are drawn from the urn and one after other. $E$ and $F$ denote respectively the events that first and second ball drawn are black.

How do I find $P(F|E)$?

I think $P(F|E) =$ the probability of occurrence of $F$ when $E$ is taken as sample space. When $E$ took place we had a black ball which should now be sample space for $F$. So, now $P(F|E)$ should be $1$. But book says $P(F|E) = 9/14$. where am I wrong.

In all questions I did in previous exercise of my book I used $P(A|B) =$ Probability of occurrence of $A$ when $B$ is taken as sample space. So if I already have done event $E$, now I should have only a black ball in my hand. Now $P(F|E)$ should mean I take out next black black ball from my black ball obtained in $E$. This gives probability of $1$. Another way to find $P(F|E)$ can be by $n\left\lvert F \cap E\right\lvert / n\left\lvert E \right\lvert$ . Can you please explain using this method?

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  • $\begingroup$ After the black ball is drawn, you have $14$ balls left of which $9$ are black. $\endgroup$ – André Nicolas Jan 9 '16 at 7:58
  • $\begingroup$ But why E is not taken as sample space. $\endgroup$ – Anubhav Goel Jan 9 '16 at 8:00
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    $\begingroup$ The sample space is the set of all pairs $(x,y)$ where the first is black. Not really a useful way to think in this case. It is clear that the conditional probability the second is black given the first is cannot be $1$, there are lots of white balls left. $\endgroup$ – André Nicolas Jan 9 '16 at 8:07
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    $\begingroup$ That's not right. The problem is with respect to the balls in the urn not from you hand. That is what is explicitly stated in the problem. We would need more context regarding the previous problems to be able to determine the issue. $\endgroup$ – Em. Jan 9 '16 at 8:16
  • $\begingroup$ Welcome to math.SE! Here is a guide to some of the tricky maths markup. $\endgroup$ – Frentos Jan 9 '16 at 9:13
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The urn contained $10$ black and $5$ white balls, then two were drawn without replacement.

In all questions I did in previous exercise of my book I used $P(A\mid B)=$ Probability of occurrence of $A$ when $B$ is taken as sample space. So if I already have done event $E$, now I should have only a black ball in my hand. Now $P(F|E)$ should mean I take out next black black ball from my black ball obtained in $E$. This gives probability of $1$.

No, the second ball must be drawn from those remaining in the urn.

There are $140$ ways to obtain event $E$:   That is to draw one of the $10$ black balls then draw any of the remaining $14$ balls.   That is your sample space.   $\mathsf P(F\mid E)$ is thus the probability measure that given we have one of these $140$ equally-probable outcomes we also obtain event $F$ - that the second ball is also black.

$$\mathsf P(F\mid E) =\frac{\mathsf P(E\cap F)}{\mathsf P(E)} = \frac{90}{140}$$

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    $\begingroup$ @AnubhavGoel The probability space consists of a draw of two balls from the urn; and there is a total of $15\times 14$ ways to do this. Of these, $10\times 14$ have a black ball drawn first and any ball drawn second. That is event $E$ - that the first of two balls drawn is black. $\endgroup$ – Graham Kemp Jan 9 '16 at 9:59
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Close. Conditioning on $E$ means that I tell you on the first draw, you got a black ball. Thus in the next turn, what is the probability that you get a black? Well, there are only 14 balls left, and 9 of them are black. Thus $$P(F|E) = \frac{9}{14}$$ using the logic above. You can check by hand using Bayes' rule.

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  • $\begingroup$ My question still stands why you did not took E as sample space. $\endgroup$ – Anubhav Goel Jan 9 '16 at 8:03
  • $\begingroup$ @AnubhavGoel What do you mean by that? Can you write it out? $\endgroup$ – Em. Jan 9 '16 at 8:06
  • $\begingroup$ @AnubhavGoel: $E$ was taken as a sample space. In $E$, we've chosen one black ball, so there are $9$ black balls and $5$ white balls left for the second choice. $\endgroup$ – robjohn Jan 9 '16 at 9:06
  • $\begingroup$ @robjon: if you took E as sample space you have only 1 black ball as sample space which you took out. $\endgroup$ – Anubhav Goel Jan 9 '16 at 9:24
  • $\begingroup$ @AnubhavGoel No, the correct sample space $E$ is the one everyone is describing about the remaining balls in the urn. The sample space $E$ is not what is in your hand. $\endgroup$ – Em. Jan 9 '16 at 9:32

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