Question

There are $6$ boxes labelled $B_1,B_2,....,B_6.$ In each trial, two fair dice $D_1,D_2$ are thrown. If $D_1$ shows $j$ and $D_2$ shows $k$, then $j$ balls are put into the box $B_k$. After $n$ trials, what is the probability that $B_1$ contains at most one ball?

• A. $(\frac{5^{n-1}}{6^{n-1}})+(\frac{5^n}{6^n})(\frac{1}{6})$
• B. $(\frac{5^n}{6^n})+(\frac{5^{n-1}}{6^{n-1}})(\frac{1}{6})$
• C. $(\frac{5^n}{6^n})+n(\frac{5^{n-1}}{6^{n-1}})(\frac{1}{6})$
• D. $(\frac{5^n}{6^n})+n(\frac{5^{n-1}}{6^{n-1}})(\frac{1}{6^2})$

Answer 1

The correct option is D $(\frac{5^n}{6^n})+n(\frac{5^{n-1}}{6^{n-1}})(\frac{1}{6^2})$

Let $P\left(B_1\right)$ be the probability that $B_1$ contains atmost one ball i.e.,
$P\left(B_1\right)=P\left(B_1\text{contains 0 balls}\right)+P\left(B_1\text{contains 1 balls}\right)$
$=P\left(D_2\text{never show 1}\right)+P\left(D_2\text{show 1 once when}{D}_1\text{show 1}\right)$
$=(\frac{5}{6}{)}^n+{}^n{C}_1(\frac{5}{6}{)}^{n-1}(\frac{1}{6})(\frac{1}{6})$
Therefore, required probability$P\left(B_1\right)=(\frac{5}{6}{)}^n+n(\frac{5}{6}{)}^{n-1}(\frac{1}{6}{)}^2$