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,
[tex]\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
The total number of balls is
[tex]3+3+4=10[/tex]
Using the above formula, we get
[tex]P(Orange)=\dfrac{3}{10}[/tex]
[tex]P(Yellow)=\dfrac{3}{10}[/tex]
[tex]P(Red)=\dfrac{4}{10}[/tex]
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
[tex]P(\text{Orange and orange})=\dfrac{3}{10}\times \dfrac{3}{10}[/tex]
[tex]P(\text{Orange and orange})=\dfrac{9}{100}[/tex]
[tex]P(\text{Orange and orange})=0.09[/tex]
Therefore the probability of getting the 2 orange balls is 0.09.
2) The probability of getting the 2 yellow balls is
[tex]P(\text{yellow and yellow})=\dfrac{3}{10}\times \dfrac{3}{10}[/tex]
[tex]P(\text{yellow and yellow})=\dfrac{9}{100}[/tex]
[tex]P(\text{yellow and yellow})=0.09[/tex]
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
[tex]P(\text{Red and yellow})=\dfrac{4}{10}\times \dfrac{3}{10}[/tex]
[tex]P(\text{Red and yellow})=\dfrac{12}{100}[/tex]
[tex]P(\text{Red and yellow})=0.12[/tex]
Therefore the probability of getting the 1 red and 1 yellow ball is 0.12.
