Chapter 13.10, Problem 2BCYP

To determine

The probability of red marble and then a white marbles chosen is to be estimated.

Answer to Problem 2BCYP

The probability is $\frac{1}{6}$ .

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 first choosing red white marble is calculated as:

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

Now, one red marble is left. So there are $7$ red marbles remain.

The probability of choosing white marble is calculated as:

  $\begin{array}{l}P\left(B\right)=\frac{M_w}{M_b+M_r+M_w}\\ P\left(B\right)=\frac{5}{3+6+5}\\ P\left(B\right)=\frac{5}{14}\end{array}$

So, the probability of choosing red marble and then a white marbles is given as:

  $\begin{array}{l}P=P\left(A\right)\times P\left(B\right)\\ P=\frac{7}{15}\times \frac{5}{14}\\ P=\frac{1}{6}\end{array}$

Conclusion:

Hence, the probability of choosing red marble and then a white marbles is $P=\frac{1}{6}$ .