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If 65% of the population has black eyes, 25% have brown eyes and the remaining have blue eyes, what is the probability that a person selected at random has: i. blue eyes ii. brown or black eyes?

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Question

If 65% of the population has black eyes, 25% have brown eyes and the remaining have blue eyes, what is the probability that a person selected at random has:

  1. blue eyes 
  2. brown or black eyes?
Sum
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Solution

Given: \[ P(\text{Black}) = 65% = 0.65 \] \[ P(\text{Brown}) = 25% = 0.25 \]

(i) Blue eyes:

Remaining percentage with blue eyes: 

\[ P(\text{Blue}) = 100% - (65% + 25%) = 10% \]

\[ P(\text{Blue}) = \dfrac{10}{100} = \dfrac{1}{10} = 0.1 \]

(ii) Brown or black eyes:

\[ P(\text{Brown or Black}) = P(\text{Brown}) + P(\text{Black}) = 25% + 65% = 90% \] 

\[ P(\text{Brown or Black}) = \dfrac{90}{100} = \dfrac{9}{10} = 0.9 \]

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Chapter 16: Probability - EXERCISE 16.1 [Page 16.22]

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R.D. Sharma Mathematics [English] Class 10
Chapter 16 Probability
EXERCISE 16.1 | Q 39. | Page 16.22
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