Question

In a certain population $10%$ of the people are rich, \[5%\] are famous. and \[3%\] are rich and famous. The probability that a person picked at random from the population is either famous or rich and not both, is equal to
A. \[0.07\]
B. \[0.08\]
C. \[0.09\]
D. \[0.12\]

Answer 1

Hint: In the given question, we are to find the probability that a person is picked at random from a given population is either rich or famous and not both. To solve this, set theory and probability are applied.



Formula Used:A probability is the ratio of favourable outcomes of an event to the total number of outcomes. So, the probability lies in between 0 and 1.
The probability is calculated by,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] - favourable outcomes and \[n(S)\] - sample.
If there are two events in a sample space, then the addition theorem on probability is given by
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
According to set theory,
The union of sets is described as:
\[A\cup B=\{x/x\in A\text{ or }x\in B\}\]
The intersection of sets is described as:
\[A\cap B=\{x/x\in A\text{ and }x\in B\}\]
The symmetric difference is described as:
\[\begin{align}
  & A-B=\{x/x\in A\text{ and }x\notin B\} \\
 & B-A=\{x/x\in B\text{ and }x\notin A\} \\
\end{align}\]



Complete step by step solution:It is given that,
The probability of rich people in the population is,
\[P(R)=\dfrac{10}{100}=0.1\]
The probability of famous people in the population is,
\[P(F)=\dfrac{5}{100}=0.05\]
The probability of rich and famous people in the population is,
\[P(R\cap F)=\dfrac{3}{100}=0.03\]
By the addition theorem on probability,
\[P(A\cup B)=P(R)+P(F)-P(R\cap F)\]
Then, the required probability is
\[\begin{align}
  & =P(R\cup F)-P(R\cap F) \\
 & =P(R)+P(F)-P(R\cap F)-P(R\cap F) \\
 & =P(R)+P(F)-2P(R\cap F) \\
\end{align}\]
On substituting,
\[\begin{align}
  & =0.1+0.05-2\times 0.03 \\
 & =0.15-0.06 \\
 & =0.09 \\
\end{align}\]



Option ‘C’ is correct

Note: In this question, we are to find the probability of an event. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability.