Question

n urn contains 3 red balls, 9 green, 2 yellow, 2 orange, and 4 purple balls. Two balls aredrawn, one at a time with replacement. Find the probability of drawing a green ball and an orangeball.

Answer 1

`(9)/(100)`

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:

`P(E) = \frac{\text{Number of favorable cases}}{\text {Total number of cases}}`

Probability that a green ball is drawn at first:

`P(Green) = \frac{\text{Number of Green balls}}{\text {Total number of Balls}}`

`P(Green) = (9)/(20)`

Now, the ball is replaced , so total number of balls remain the same i.e. 20.
`P(Orange) = \frac{\text{Number of Orange balls}}{\text {Total number of Balls}}`

`P(Orange) = (2)/(20) = (1)/(10)`

`P(Green\ then\ orange) = P(Green) * P(Orange)\\\Rightarrow P(Green\ then\ orange) = (9)/(10) * (1)/(10)\\\Rightarrow P(Green\ then\ orange) = \bold{ (9)/(100) }`