Chapter 12.8, Problem 3STP

For Exercises 1-5, choose the correct letter.

You choose a marble at random from a bag containing 3 orange marbles, 4 green marbles, and 3 yellow marbles. You replace the marble and then pick a second marble. What is the probability that you will pick a green marble both times?

1. $\frac{6}{50}$

B. $\frac{2}{15}$

C. $\frac{4}{25}$

D. $\frac{8}{45}$

To determine

Find the probability :

P(both green)

Answer to Problem 3STP

C

Explanation of Solution

Given:

You choose a marble at random from a bag containing 3 orange marbles, 4 green marbles , and 3 yellow marbles.You replace the marble and then pick a second marble .

Formula Used:

  $P(E)=\frac{\text{Favourable outcomes of E}}{\text{total number of outcomes}}$

  $P\left(AandB\right)=P\left(A\right)\cdot P\left(B\right)$

Calculation:

Number of orange marbles = 3

Number of green marbles = 4

Number of yellow marbles = 3

Total marbles = 10

Number of favorable outcomes of first green marble = 4

Number of favorable outcomes of second green marble = 4 (since the marble is replaced after first turn)

  $P(\text{first green})=\frac{\text{favourable outcomes }}{total\text{ number of outcomes}}=\frac{4}{10}=\frac{2}{5}$

  $P(\sec ond\text{ green})=\frac{\text{favourable outcomes }}{total\text{ number of outcomes}}=\frac{4}{10}=\frac{2}{5}$

  $\begin{array}{l}P(both\text{ green})=P(first\text{ green})\cdot P(\sec ond\text{ green)}\\ =\frac{2}{5}\cdot \frac{2}{5}=\frac{4}{25}\end{array}$

Hence , the correct option is C.