In a town of 6000 people, 1200 are over 50 years old, and 2000 are females. It is known that 30% of the females are over 50 years. What is the probability that a randomly chosen individual from the town is either female or over 50 years?
let A denote the event that the chosen individual is female and B denote the event that the chosen individual is over 50 years old.
Given : Town consists of 6000 people, 1200 are over 50 years old, and 2000 are females
To find : Probability that a randomly chosen individual from the town is either female or over 50 years $=P(A$ or $B)$
The formula used : Probability =
$P(A$ or $B)=P(A)+P(B)-P(A$ and $B)$
For the event $A$,
There are 2000 females present in a town of 6000 people
Favourable number of outcomes $=2000$
Total number of outcomes $=6000$
$P(A)=\frac{2000}{6000}=\frac{1}{3}$
For the event B,
There are 1200 are over 50 years of age in a town of 6000 people
Favourable number of outcomes $=1200$
Total number of outcomes $=6000$
$P(A)=\frac{1200}{6000}=\frac{1}{5}$
$30 \%$ of the females are over 50 years
For the event $A$ and $B$,
$\frac{30}{100} \times 2000=600$
females are over 50 years of age
Favourable number of outcomes = 600
$P(A$ and $B)=\frac{600}{6000}=\frac{1}{10}$
$P(A$ or $B)=P(A)+P(B)-P(A$ and $B)$
$P(A$ or $B)=\frac{1}{3}+\frac{1}{5}-\frac{1}{10}$
$P(A$ or $B)=\frac{10+6-3}{30}=\frac{13}{30}$
$P(A$ or $B)=\frac{13}{30}$
The probability that a randomly chosen individual from the town is either female or over
50 years $=P(A$ or $B)=\frac{13}{30}$
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