I have some trouble understanding the conditional probability of the following question.. Do pardon me even though it is a simple question..

Two fair dice are rolled and the sum on the faces is calculated. Find the probability : Pr(Sum of two dice is 8 | we know that the number on both dice is even)

This is my working

Let E be the event that the Sum of two dice is 8. $n(E) = {(2,6),(3,5),(4,4),(5,3),(6,2)}, P(E) = 5/36 $

Let F be the event that the number on both dice is even $n(F) = {(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)}, P(F) = 9/36 $

$P(E|F) = (5/36)/(9/36) = 5/9 $

However, in my textbook, while P(F) is the same answer as mine but P(E) is written as 3/36 and hence the overall answer is 1/3 I am unable to get my head around the part for P(E), other than 3/36 is obtained due to ${(2,6),(4,4),(6,2)}$ as these three are the even numbers..

Am I correct to assume as such?

  • 1
    $\begingroup$ You enumerated all possible outcomes given both die are even...only three of the $9$, $\{(2, 6), (4, 4), (6, 2)\}$, result in a sum of $8$, hence it's $\frac{3}{9} = \frac{1}{3}$. $\endgroup$ – Jared Apr 20 '16 at 4:11
  • $\begingroup$ @Jared okay.. I think I am having issues interpreting when given a conditional probability. This question is a good example of it though.. Many thanks $\endgroup$ – dissidia Apr 20 '16 at 4:34

The conditional probability $\mathbb{P}(E|F)$ is not equal to $\frac{\mathbb{P}(E)}{\mathbb{P}(F)}$, but rather to $\frac{\mathbb{P}(E\cap F)}{\mathbb{P}(F)}$.

There are three outcomes in $E\cap F$, namely $(2,6),(4,4)$, and $(6,2)$, hence the conditional probability is $$ \frac{\frac{3}{36}}{\frac{9}{36}}=\frac{1}{3}$$

  • $\begingroup$ My bad on the 'signs'. Will take note of that. I think I am still having issues understand when given a conditional prob. So suppose if now the question is stated something like Pr(Sum of two dice is 8 | we know that the number on both dice is <= 10), it will then be (5/36)/(33/36)? $\endgroup$ – dissidia Apr 20 '16 at 4:32
  • $\begingroup$ I'm not sure I understand, aren't the numbers on standard dice always at most 6? $\endgroup$ – carmichael561 Apr 20 '16 at 4:36
  • $\begingroup$ @carmichael561 Unless dissidia meant "... given that the sum is at most 10", which would indeed have a probability of $5/33$. $\endgroup$ – Graham Kemp Apr 20 '16 at 4:40
  • $\begingroup$ Perhaps that is it. $\endgroup$ – carmichael561 Apr 20 '16 at 4:41
  • 1
    $\begingroup$ @dissidia In this case you have $E\subset G$ so $E\cap G=E$ $$\Pr(E\mid G) = \dfrac{\Pr(E\cap G)}{\Pr(G)} = \dfrac{\Pr(E)}{\Pr(G)} = \dfrac{5}{33}$$ $\endgroup$ – Graham Kemp Apr 20 '16 at 4:45

You wish to know the probability that the sum is eight ($E$) given that the two dice are even ($F$). $\Pr(E\mid F)$

As you stated $E$ is the event set: $\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$ and $F$ is the event set $\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}$

However, we want only want the conditional probability of $E$ when $F$.

Now the event of both dice being even and the sum equaling eight is: $F\cap E = \{(2,6) (4,4), (6,2)\}$

Since all outcomes of this experiment are equally likely $\Pr(E\mid F) = \dfrac {\lvert F\cap E\rvert}{\lvert F\rvert} = \dfrac{3}{9}$


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