A jar contains 3 orange balls, 3 yellow balls, and 4 red balls. Providing the ball drawn first is replaced before the second is drawn. What is the probability of selecting: 1. 2 orange balls
2. 2 yellow balls
3. 1 red and 1 yellow ball

Given:

Number of orange balls = 3

Number of yellow balls = 3

Number of red balls = 4

To find:

The probability of:

1) 2 orange balls

2) 2 yellow balls

3) 1 red and 1 yellow ball

Solution:

We know that,

The total number of balls is

Using the above formula, we get

The ball drawn first is replaced before the second is drawn. So, the probabilities remains unchanged.

1) The probability of getting the 2 orange balls is

Therefore the probability of getting the 2 orange balls is 0.09.

2) The probability of getting the 2 yellow balls is

Therefore the probability of getting the 2 yellow balls is 0.09.

3) The probability of getting the 1 red and 1 yellow ball is

Therefore the probability of getting the 1 red and 1 yellow ball is 0.12.

4 + 5 7/9= 9 7/9

step-by-step explanation:

c 1 hour

step-by-step explanation:

m = 2e

m = 2 so 2 = 2e

1 = e