# conditional probability question

• Feb 24th 2013, 09:27 AM
anthonye
conditional probability question
Hi;
the question is two dice are thrown and there product is recorded,
given that one die lands on a 2 find the probability of the dice is exactly 6?

I get 1/6 = 0.1667 the reason I get this is because the other dice needs to land on a 3.

is this correct?
• Feb 24th 2013, 09:32 AM
BobSacamano
Re: conditional probability question
assuming the question is 'find the probability that the product of the dice is exactly 6' then yes correct
• Feb 24th 2013, 09:43 AM
Re: conditional probability question
If you wanted to show it formally (not necessary however)

Let A and B be events with A = "The first die is 2" and B = "The product of the dice is 6", then

$\displaystyle P(B|A) = \frac{P(A and B)}{P(B)}$

As you said, the other dice needs to be a 3 and hence $\displaystyle P(A and B) = \frac{1}{36}$ and $\displaystyle P(B) = \frac{1}{6}$ and hence $\displaystyle P(B|A) = \frac{1}{6}$
• Feb 24th 2013, 09:46 AM
anthonye
Re: conditional probability question
Thank you;
Does the probability stay the same if its the second dice that shows a 2?
• Feb 24th 2013, 10:47 AM
Plato
Re: conditional probability question
Quote:

Originally Posted by anthonye
the question is two dice are thrown and there product is recorded, given that one die lands on a 2 find the probability of the dice is exactly 6?

Perhaps I have misread the question. Is it, "find the probability that the product of the dice is exactly 6, given that one die lands on a 2"?

If that is indeed the question then the answer is $\displaystyle \frac{2}{11}$.
Thinking in terms of pairs, there are eleven pair that contain a two. Of those, only two pair also contain a three.

If $\displaystyle A$ is the event that the product is six and $\displaystyle B$ is the event that one die lands on a 2, then
$\displaystyle \mathcal{P}(A|B)=\frac{\mathcal{P}(A\cap B)}{\mathcal{P}(B)}=\frac{2/36}{11/36}=\frac{2}{11}$
• Feb 25th 2013, 03:19 AM
anthonye
Re: conditional probability question
Thank you Plato thats exactly what I needed to know.