A simple conditional probability problem Assume that two fair dice are rolled one at a time. Given that the sum of the two numbers that occured was at least $7$, compute the probability that it was equal to $7$.
I tried computing the probability as follows:
$$P \left( X+Y=7| X+Y \geq 7 \right)=\frac{P \left(X+Y=7 \right)}{P \left(X+Y \geq 7 \right)}$$
And then I examined all different pairs whose sum is $7$ and then those whose sum is greater or equal to $7$. I am getting $3/12$ but the right answer is $6/21$. Could you please help me understand what I am doing wrong? 
Thank you.
 A: $P(X+Y=7) = 1/6$
By Symmetry
$P(X+Y>7) = P(X+Y<7) = (1 - 1/6)/2$
$P(X+Y>=7) = P(X+Y>7)+P(X+Y=7) = 5/12 + 1/6$
$P( X+Y=7| X+Y >= 7) = \frac{1/6}{5/12+1/6} = 6/21$
A: The total number of possible outcomes when rolling two fair dice is $6^2=36$. 
Examining the denominator of your conditional probability, it is easier to deal with the case when the sum of the two dice ($X+Y$) is less than $7$, and subtract this result from $36$. 
Now, there are $5$ outcomes (pairs ($5$,$1$),($4$,$2$),($3$,$3$),($2$,$4$),($1$,$5$)) for the sum to be $6$, $4$ outcomes for the sum to be $5$, and so on, until the minimum value (which is $2$, where each die shows a $1$). 
So the total number of outcomes where $X+Y < 7$ is $5+4+3+2+1=15$. 
As a result the total number of outcomes where $X+Y \geq 7$ is $36-15=21$, leading to
$$P(X+Y\geq 7)=\frac{21}{36}$$
Next, we count the number of total outcomes where $X+Y=7$, which is $6$ (($6$,$1$)($5$,$2$),($4$,$3$),($3$,$4$),($2$,$5$),($1$,$6$)), so that
$$P(X+Y=7)=\frac{6}{36}$$
We thus obtain
$$P(X+Y=7|X+Y\geq 7)=\frac{(P(X+Y=7) \cap (X+Y\geq7))}{P(X+Y\geq7)}=\frac{P(X+Y=7)}{P(X+Y\geq7)}\\=\frac{6}{36}\times\frac{36}{21}=\frac{6}{21}=\frac{2}{7}$$ 
A: You're mistake lies in the equation you used for calculating. The LHS is not equal to the RHS.
I believe the best approach to calculating the conditional probability in this instance would be to use Bayes' Theorem. 
Theorem: Pr[A|B] = (Pr[B|A]*Pr[A])/Pr[B]
So here, Pr[B|A]=Pr[X+Y>=7|X+Y=7]=1,
Pr[A]=Pr[X+Y=7]=6/36
Pr[B]=Pr[X+Y>=7]=21/36
Hence, answer is 6/21.
A: How many ordered pairs of rolls give you a number over $6$? How many give you exactly  $7$?
