Chapter 12.8, Problem 25P

Yon choose a marble at random from a bag containing 3 blue marbles, 5 red marbles, and 2 green marbles. You replace the marble and then choose again. Find each probability.

P(both green)

To determine

Find the probability :

P(both green)

Answer to Problem 25P

  $\frac{1}{25}$

Explanation of Solution

Given:

You choose a marble at random from a bag containing 3 blue marbles , 5 red marbles , and 2 green marbles.You replace the marble and then choose again.

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:

Blue marbles = 3

Red marbles = 5

Green marbles = 2

Total number of marbles = 10

Number of favorable outcomes of green marbles = 2

  $P(\text{first green})=\frac{\text{favourable outcomes }}{total\text{ number of outcomes}}=\frac{2}{10}=\frac{1}{5}$

  $P(\sec ond\text{ green})=\frac{\text{favourable outcomes }}{total\text{ number of outcomes}}=\frac{2}{10}=\frac{1}{5}$

Since , the marbles are replaced , both the events are independent.

  $\begin{array}{l}P(2\text{ green})=P(first\text{ green})\cdot P(\sec ond\text{ green)}\\ =\frac{1}{5}\cdot \frac{1}{5}=\frac{1}{25}\end{array}$