There are 6 boxes labelled B_1, B_2, ldots, B | vedclass

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(D) Let $X_i$ be the number of balls added to box $B_1$ in the $i$-th trial. In each trial,$D_1$ shows $j \in \{1, 2, 3, 4, 5, 6\}$ and $D_2$ shows $k

Vedclass · Accepted Answer

$\left(\frac{5^n}{6^n}\right)+n\left(\frac{5^{n-1}}{6^{n-1}}\right)\left(\frac{1}{6^2}\right)$

Vedclass · Answer

(D) Let $X_i$ be the number of balls added to box $B_1$ in the $i$-th trial. In each trial,$D_1$ shows $j \in \{1, 2, 3, 4, 5, 6\}$ and $D_2$ shows $k \in \{1, 2, 3, 4, 5, 6\}$. Box $B_1$ receives $j$ balls if $k=1$,and $0$ balls if $k 
eq 1$. For $B_1$ to contain at most one ball after $n$ trials,either zero balls are added in all $n$ trials,or exactly one ball is added in one trial and zero in the remaining $n-1$ trials. Case $1$: Zero balls are added in all $n$ trials. This happens if $k 
eq 1$ in every trial. The probability is $(\frac{5}{6})^n$. Case $2$: Exactly one ball is added in one trial and zero in the others. This happens if in one trial $j=1$ and $k=1$ (probability $\frac{1}{6} 	imes \frac{1}{6} = \frac{1}{36}$),and in the other $n-1$ trials $k 
eq 1$ (probability $(\frac{5}{6})^{n-1}$). Since there are $n$ choices for the trial in which the ball is added,the probability is $n 	imes \frac{1}{36} 	imes (\frac{5}{6})^{n-1} = n 	imes \frac{5^{n-1}}{6^{n-1}} 	imes \frac{1}{6^2}$. Total probability = $(\frac{5}{6})^n + n 	imes \frac{5^{n-1}}{6^{n-1}} 	imes \frac{1}{6^2}$.

There are $6$ boxes labelled $B_1, B_2, \ldots, B_6$. 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?

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