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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?

User Harpax
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By definition of conditional probability,


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


=\frac{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)=\frac{\frac1{18}}{\frac16}=\frac13

User SuperFamousGuy
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