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…

Question

asked Sep 24, 2022 83.1k views

1 Answer

Ricardo Peres · Answer 1 · 2022-10-01T06:54:08+0000

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,

$\text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

The total number of balls is

3+3+4=10

Using the above formula, we get

P(Orange)=(3)/(10)

P(Yellow)=(3)/(10)

P(Red)=(4)/(10)

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

$P(\text{Orange and orange})=(3)/(10)* (3)/(10)$

$P(\text{Orange and orange})=(9)/(100)$

$P(\text{Orange and orange})=0.09$

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

2) The probability of getting the 2 yellow balls is

$P(\text{yellow and yellow})=(3)/(10)* (3)/(10)$

$P(\text{yellow and yellow})=(9)/(100)$

$P(\text{yellow and yellow})=0.09$

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

$P(\text{Red and yellow})=(4)/(10)* (3)/(10)$

$P(\text{Red and yellow})=(12)/(100)$

$P(\text{Red and yellow})=0.12$

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