Chapter 13.10, Problem 2ACYP

To determine

The probability of two white marbles chosen is to be estimated.

Answer to Problem 2ACYP

The probability of two white marbles chosen is $\frac{2}{21}$ .

Explanation of Solution

Given:

  $\begin{array}{l}{M}_b=3\\ M_r=7\\ M_w=5\end{array}$

Concept Used:

Probability of an event is calculated as:

  $P=\frac{n\left(E\right)}{n\left(S\right)}$

Where,

The probability of an event: $P$

The number of favourable outcomes: $n\left(E\right)$

The total number of events in the sample space: $n\left(S\right)$

Calculation:

The probability of choosing first white marble is calculated as:

  $\begin{array}{l}P\left(A\right)=\frac{M_w}{M_b+M_r+M_w}\\ P\left(A\right)=\frac{5}{3+7+5}\\ P\left(A\right)=\frac{5}{15}\\ P\left(A\right)=\frac{1}{3}\end{array}$

Now, only four marbles of white colour left.

The probability of choosing second white marble is calculated as:

  $\begin{array}{l}P\left(B\right)=\frac{M_w}{M_b+M_r+M_w}\\ P\left(B\right)=\frac{4}{3+7+4}\\ P\left(B\right)=\frac{4}{14}\\ P\left(B\right)=\frac{2}{7}\end{array}$

So, the probability of choosing two white marbles is given as:

  $\begin{array}{l}P=P\left(A\right)\times P\left(B\right)\\ P=\frac{1}{3}\times \frac{2}{7}\\ P=\frac{2}{21}\end{array}$

Conclusion:

Hence, the probability of choosing two white marbles is $P=\frac{2}{21}$ .