Question

If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is $i$? Compute for all values of $i$between $2$and$12$

Short answer

$\begin{array}{r}P(A\mid i=2)=0\\ P(A\mid i=3)=0\\ P(A\mid i=4)=0\\ P(A\mid i=5)=0\\ P(A\mid i=6)=0\\ P(A\mid i=7)=\frac{1}{6}\\ P(A\mid i=8)=\frac{1}{5}\\ P(A\mid i=9)=\frac{1}{4}\\ P(A\mid i=10)=\frac{1}{3}\\ P(A\mid i=11)=\frac{1}{2}\\ P(A\mid i=12)=1\end{array}$

Step-by-step solution

Given Information

The number on first dice= $6$

Sum of both dice=$i$.

Calculation.

When i = $2,3,4,5,6$

Consider that $A$is the event that first lands in $6$.

$\begin{array}{r}P(A\mid i=2)=0\\ P(A\mid i=3)=0\\ P(A\mid i=4)=0\\ P(A\mid i=5)=0\\ P(A\mid i=6)=0.\end{array}$

$\begin{array}{r}P(A\mid i=7)=\frac{1}{6}\\ \end{array}$

$\begin{array}{r}\\ P(A\mid i=8)\text{, that is}i=8\text{occurs when}(2,6)(6,2)(5,3)(3,5)(4,4)\end{array}$

Thus,

$P(A\mid i=9)\text{, that is}i=9\text{occurs when}(3,6)(6,3)(5,4)(4,5)$

$\begin{array}{r}P(A\mid i=8)=\frac{1}{5}\\ P(A\mid i=9),\text{that}\\ \text{Thus,}\\ P(A\mid i=9)=\frac{1}{4}\end{array}$

$P(A\mid i=10)$that is $i=10$occurs when $(4,6)(6,4)(5,5)$

Thus,

$P(A\mid i=10)=\frac{1}{3}$

$P(A\mid i=11)$that is $i=11$occurs when$(5,6)(6,5)$

$P(A\mid i=11)=\frac{1}{2}$

$P(A\mid i=12)$, that is $i=12$occurs when $(6,6)$

Thus,

$P(A\mid i=12)=1$