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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 13. The probability of yellow ball is 15 What is

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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`. What is the probability of blackball?

Sum
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Solution

Let event G = A Green ball is sleeted from the bag.

Given: P(G) = `1/3`
Event B = A Blackball is selected from the bag.
To find P(B).
Event Y = A Yellow ball is selected from the bag.
Given: P(Y) =  `1/5`.

Since the events are mutually exclusive and exhaustive,
P(G) + P(B) + P(Y) = 1
∴ `1/3 + "P"("B") + 1/5` = 1

∴ P(B) + `8/15` = 1

∴ P(B) = `1 - 8/15 = 7/15`
∴ Probability of a black ball selected is `7/15`.

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Chapter 7: Probability - Exercise 7.2 [Page 102]
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