Answer:
[tex]\frac{9}{100}[/tex]
Step-by-step explanation:
Given:
Number of red balls, n(R) = 3
Number of green balls, n(G) = 9
Number of yellow balls, n(Y) = 2
Number of orange balls, n(O) = 2
Number of purple balls, n(P) = 4
Two balls are drawn one at a time with replacement.
To find:
Probability of drawing a green ball and an orange ball ?
Solution:
Total number of balls, n(Total) = 3 + 9 + 2 + 2 + 4 = 20
Formula for probability of an event E is given as:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
Probability that a green ball is drawn at first:
[tex]P(Green) = \dfrac{\text{Number of Green balls}}{\text {Total number of Balls}}[/tex]
[tex]P(Green) = \dfrac{9}{20}[/tex]
Now, the ball is replaced , so total number of balls remain the same i.e. 20.[tex]P(Orange) = \dfrac{\text{Number of Orange balls}}{\text {Total number of Balls}}[/tex]
[tex]P(Orange) = \dfrac{2}{20} = \dfrac{1}{10}[/tex]
[tex]P(Green\ then\ orange) = P(Green) \times P(Orange)\\\Rightarrow P(Green\ then\ orange) = \dfrac{9}{10} \times \dfrac{1}{10}\\\Rightarrow P(Green\ then\ orange) = \bold{ \dfrac{9}{100} }[/tex]
