Question

A bag contains 15 balls of three different colours: Green, Black and Yellow. A ball is drawn at random from the bag. The probability of green ball is `1/3`. The probability of yellow ball is `1/5`. How many balls are green, black, and yellow?

Answer 1

n(S) = 15

Let, Green, Black and Yellow balls be G, B, and Y, respectively.

P(G) = `1/3`   ...(Given)

P(G) = `"n(G)"/"n(S)"`

`1/3 = "n(G)"/15`

`15/3` = n(G)

∴ n(G) = 5

P(Y) = `1/5`   ...(Given)

P(Y) = `"n(Y)"/"n(S)"`

`1/5 = "n(Y)"/15`

`15/5` = n(Y)

∴ n(Y) = 3

Total balls = Green balls + Yellow balls + Black balls

15 + 5 + 3 + Black balls

15 = 8 + Black balls

15 − 8 = Black balls

Black balls = 7

∴ n(B) = 7

There are 5 Green balls 7 Black balls and 3 Yellow balls in the bag.