Question

I roll a fair die twice and obtain two numbers X1= result of the first roll and X2= result of the second roll. Given that I know X1+X2=7, what is the probability that X1=4 or X2=4?

Answer 1

By definition of conditional probability,

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{P((X_1=4\text{ or }X_2=4)\text{ and }X_1+X_2=7)}{P(X_1+X_2=7)}$

$=\dfrac{P((X_1=4\text{ and }X_1+X_2=7)\text{ or }(X_2=4\text{ and }X_1+X_2=7))}{P(X_1+X_2=7)}$

Assuming a standard 6-sided fair die,

• if $X_1=4$, then $X_1+X_2=7$ means $X_2=3$; otherwise,
• if $X_2=4$, then $X_1=3$.

Both outcomes are mutually exclusive with probability $\frac1{36}$ each, hence total probability $\frac2{36}=\frac1{18}$.

Of the 36 possible outcomes, there are 6 ways to sum the integers 1-6 to get 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and so a sum of 7 occurs $\frac6{36}=\frac16$ of the time.

Then the probability we want is

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13$