Question

In India, if $60\%$ population likes tea, $10\%$ likes coffee and $5\%$ likes both. A person is selected at random, then the probability that he(she) does not like both is

• A. $35\%$
• B. $20\%$
• C. $25\%$
• D. $30\%$

Answer 1

The correct option is A $35\%$

Given : $P\left(T\right)=\frac{60}{100}=\frac{6}{10},$
$P\left(C\right)=\frac{10}{100}=\frac{1}{10},$
$P(T\cap C)=\frac{5}{100}$
So, Required probability
$=P(T^c\cap C^c)=P(T\cup C{)}^c=1-P(T\cup C)=1-[\frac{6}{10}+\frac{1}{10}-\frac{5}{100}]=1-\frac{65}{100}=35\%$