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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?
Solution
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.
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