Question

In a town of 6000 people 1200 are over 50 years old and 2000 are female. It is known that 30% of the females are over 50 years. What is the probability that a random chosen individual from the town either female or over 50 years ?

• A. $\frac{5}{17}$
• B. $\frac{13}{30}$
• C. $\frac{17}{55}$
• D. $\frac{3}{5}$

Answer 1

Answer: (B)

$\frac{13}{30}$

Consider the following events :

A = A randomly chosen individual is a female

B = A randomly chosen individual is over 50 years old

$P\left(A\right)=\frac{2000}{6000}=\frac{1}{3},P\left(B\right)=\frac{1200}{6000}=\frac{1}{5}$

and, $P(A\cap B)=$ P( An individual is a female over 50 years old ) = $\frac{30\%of2000}{6000}=\frac{600}{6000}=\frac{1}{10}$

Required Probability = $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{1}{3}+\frac{1}{5}-\frac{1}{10}=\frac{13}{30}$