Question

Mareena rolled a fair die twice. The result of the first roll is X₁ and the result of the second roll is X₂.

If X₁ + X₂ = 7, then the probability that X₁ = 4 or X₂ = 4 is around _____

4 Decimal Places

Answer 1

By the definition of conditional probability,

$P(X_1 = 4 or X_2 = 4 | X_1+X_2=7) = \dfrac{P(X_1 = 4 \hspace{1mm} or \hspace{1mm} \hspace{1mm}X_2 = 4 \hspace{1mm} and \hspace{1mm}P(X_1+X_2=7)}{P(X_1+X_2 = 7)}$

$= \dfrac{P((X_1 = 4\hspace{1mm} and\hspace{1mm} X_1+X_2 = 7) or (X_2=4 \hspace{1mm}and\hspace{1mm} X_1+X_2=7)) }{P(X_1+X_2 = 7)}$

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 $\dfrac{1}{36}$ each, Hence total probability

$= \dfrac{1}{36}+ \dfrac{1}{36} = \dfrac{1}{18}$

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 $\dfrac{6}{36} = \dfrac{1}{6}$ of the time.

Then the probability we want is:

$P(X_1 =4 or X_2 | X_1+X_2 = 7) = \dfrac{\dfrac{1}{18}}{\dfrac{1}{6}} = \dfrac{1}{3}$