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3.2

Expert-verifiedFound in: Page 97

Book edition
9th

Author(s)
Sheldon M. Ross

Pages
432 pages

ISBN
9780321794772

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 $12$

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

The number on first dice= $6$

Sum of both dice= $i$.

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,

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

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