You roll two six-sided dice, one then the other.

Let A = "The first die is a 1 or 4" B = "Sum of the two dice is strictly greater than 7"

What is the $P(B \mid A)$? Give your answer out to two decimal places.

so I worked out that the probability of is :

A is 1/3 B is 5/12

So P(B given A) would be:

number that's 4 and greater than 7/number that first roll is 4

but that would give me 3 which is wrong. So what is the formula for a conditional like this?


Given $A$, the event $B$ requires two independent events: the first die must be a $4$ and the second must be one of $\{4,5,6\}$. The second has nothing to do with $A$ and has probability $\frac 12$ (conditioned on $A$ or not). The first, however, has a lot to do with $A$ so let's analyze it. Given $A$, there is an equal chance that the first die is $1$ or $4$ (as these are equally likely rolls of a die) hence $$P(first\; die\; is \;4|A)=\frac 12$$

Thus the answer is $$\frac 12 \times \frac 12=\frac 14$$


The formula for the conditional probability that event $B$ occurs given that event $A$ has occurred is $$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$ You found that $P(A) = \frac{1}{3}$. Moreover, you discovered that there were three ways for event $B$ to occur given that event $A$ had already occurred. The number of events in the sample space is $36$ since there are $6$ possible outcomes for the second roll for each of the six possible outcomes for the first roll. Hence, $P(A \cap B) = \frac{3}{36} = \frac{1}{12}$. Thus, $$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{4}$$

  • $\begingroup$ @drhab Indeed, I did. Thank you. $\endgroup$ – N. F. Taussig Jan 31 '16 at 16:32

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