3.2 - A First Course in Probability

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 $12$

Step-by-Step Solution

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 1 : Given Information

The number on first dice= $6$

Sum of both dice= $i$.

Step 2 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, 

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