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A bag contains 4 white, 6 red and 5 blue balls out of which one ball is drawn at random. The probability that it is neither white nor red is

A) \( \Large \frac{5}{15} \)
B) \( \Large \frac{4}{15} \)
C) \( \Large \frac{6}{15} \)
D) \( \Large \frac{3}{15} \)

Correct Answer:

A) \( \Large \frac{5}{15} \)

Description for Correct answer:
There are 4 white, 6 red and 5 blue balls.

Total number of balls = \( \Large n \left(S\right) \) = 15.

Number of events to draw white ball = \( \Large n \left(E_{1}\right) \) = 4

Number of events to draw red ball = \( \Large n \left(E_{2}\right) \) = 6

Number of events to draw blue ball = \( \Large n \left(E_{3}\right) \) = 5

Therefore, \( \Large P \left(E_{1}\right)=\frac{4}{15};\ P \left(E_{2}\right)=\frac{6}{15};\ P \left(E_{3}\right)=\frac{5}{15} \)

Therefore, \( \Large P \left(E_{1}\right)+P \left(E_{2}\right) = \frac{4}{15}+\frac{6}{15} = \frac{10}{15} \)

Probability it is neither white nor red is = \( \Large 1 - \frac{10}{15} = \frac{5}{15} \)

Part of solved Statistics questions and answers : >> Elementary Mathematics >> Statistics

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