Question

An urn contains $5$white and $10$black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on $3$if all the balls selected are white?

Short answer

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

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

Step-by-step solution

Given Information

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:$10$.

Solution of the Problem

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: $P\left(R_i\right)=\frac{1}{6}$

$P\left(W\mid R_1\right)=\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)}{\left(\begin{array}{c}15\\ 1\end{array}\right)}\Rightarrow \frac{5}{15}\Rightarrow \frac{1}{3}$

$P\left(W\mid R_2\right)=\frac{\left(\begin{array}{c}5\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 2\end{array}\right)}\Rightarrow \frac{10}{105}\Rightarrow \frac{2}{21}$

Computation of the Value

Simplifying the equation,

$P\left(W\mid R_3\right)=\frac{\left(\begin{array}{c}5\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}\Rightarrow \frac{10}{455}\Rightarrow \frac{2}{91}$

$P\left(W\mid R_4\right)=\frac{\left(\begin{array}{c}5\\ 4\end{array}\right)}{\left(\begin{array}{c}15\\ 4\end{array}\right)}\Rightarrow \frac{5}{1365}\Rightarrow \frac{1}{273}$

$P\left(W\mid R_5\right)=\frac{\left(\begin{array}{c}5\\ 5\end{array}\right)}{\left(\begin{array}{c}15\\ 5\end{array}\right)}\Rightarrow \frac{1}{3003}$

We get,

$P\left(W\mid R_6\right)=0.$

Computation of the Probability

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

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

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

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

$=\frac{1365}{18018}$

We get,

$\cong 0.0758.$

Computation of the Conditional Probability

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

$P\left(R_3\mid W\right)=\frac{P\left(W\mid R_3\right)P\left(R_3\right)}{P\left(W\right)}$

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

$=\frac{0.021978\times 0.1667}{0.0758}$

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

$\cong 0.0483.$

Final Answer

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 $0.0483.$