By definition of conditional probability,
[tex]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)}[/tex]
[tex]=\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)}[/tex]
Assuming a standard 6-sided fair die,
- if [tex]X_1=4[/tex], then [tex]X_1+X_2=7[/tex] means [tex]X_2=3[/tex]; otherwise,
- if [tex]X_2=4[/tex], then [tex]X_1=3[/tex].
Both outcomes are mutually exclusive with probability [tex]\frac1{36}[/tex] each, hence total probability [tex]\frac2{36}=\frac1{18}[/tex].
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 [tex]\frac6{36}=\frac16[/tex] of the time.
Then the probability we want is
[tex]P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13[/tex]
