91Ó°ÊÓ

The number of balls in a particular urn is: $15$.

Out of $15$balls, the number of white balls is: $5$.

Out of $15$ balls, the number of black balls is:$10$.

Let $R_i$denote the event of getting an$i$while rolling the die (for $i=1,2,3,4,5,6$)

Let $W$denote the event that all the selected balls are white.

The probability that $R_i$is: $P\left(R_i\right)=\frac{1}{6}$

$P\left(Wâˆ\pounds R_1\right)=\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)}{\left(\begin{array}{c}15\\ 1\end{array}\right)}⇒\frac{5}{15}⇒\frac{1}{3}$

$P\left(Wâˆ\pounds R_2\right)=\frac{\left(\begin{array}{c}5\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 2\end{array}\right)}⇒\frac{10}{105}⇒\frac{2}{21}$

Simplifying the equation,

$P\left(Wâˆ\pounds R_3\right)=\frac{\left(\begin{array}{c}5\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}⇒\frac{10}{455}⇒\frac{2}{91}$

$P\left(Wâˆ\pounds R_4\right)=\frac{\left(\begin{array}{c}5\\ 4\end{array}\right)}{\left(\begin{array}{c}15\\ 4\end{array}\right)}⇒\frac{5}{1365}⇒\frac{1}{273}$

$P\left(Wâˆ\pounds R_5\right)=\frac{\left(\begin{array}{c}5\\ 5\end{array}\right)}{\left(\begin{array}{c}15\\ 5\end{array}\right)}⇒\frac{1}{3003}$

We get,

$P\left(Wâˆ\pounds R_6\right)=0.$

The probability that all of the balls selected are white is,$P\left(W\right)=P\left(Wâˆ\pounds R_i\right)\left(\begin{array}{l}P\left(Wâˆ\pounds R_1\right)+P\left(Wâˆ\pounds R_2\right)+P\left(Wâˆ\pounds R_3\right)\\ +P\left(Wâˆ\pounds R_4\right)+P\left(Wâˆ\pounds R_5\right)+P\left(Wâˆ\pounds R_6\right)\end{array}\right)$

$=\frac{1}{6}\left(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0\right)$

$=\frac{1}{6}\left(\frac{1001+286+66+11+1}{3003}\right)$

We get,$=\frac{1}{6}\left(\frac{1365}{3003}\right)$

$=\frac{1365}{18018}$

We get,

$â‰\dots 0.0758.$

The conditional probability that the die landed on$3$ if all the balls selected are white is,

$P\left(R_3âˆ\pounds W\right)=\frac{P\left(Wâˆ\pounds R_3\right)P\left(R_3\right)}{P\left(W\right)}$

$=\frac{\left(\frac{2}{91}\right)\left(\frac{1}{6}\right)}{(0.0758)}$

$=\frac{0.021978×0.1667}{0.0758}$

We get$=\frac{0.00366}{0.0758}$

$â‰\dots 0.0483.$

The probability that all of the balls selected are white is$0.0758.$

The conditional probability that the die landed on 3 if all the balls selected are white is $0.0483.$