In a certain population,10% of the people are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $R$ be the event that a person is rich and $F$ be the event that a person is famous.Given: $P(R) = 10\% = 0.1$,$P(F) = 5\% = 0.05$,and $P(R \c

Vedclass · Accepted Answer

$0.09$

Vedclass · Answer

(C) Let $R$ be the event that a person is rich and $F$ be the event that a person is famous.Given: $P(R) = 10\% = 0.1$,$P(F) = 5\% = 0.05$,and $P(R \cap F) = 3\% = 0.03$.We need to find the probability that a person is either rich or famous but not both,which is given by the symmetric difference $P(R \Delta F) = P(R \cup F) - P(R \cap F)$.Alternatively,$P(R \Delta F) = P(R) + P(F) - 2P(R \cap F)$.Substituting the values:$P(R \Delta F) = 0.1 + 0.05 - 2(0.03)$$P(R \Delta F) = 0.15 - 0.06 = 0.09$.

In a certain population,$10\%$ of the people are rich,$5\%$ are famous,and $3\%$ are both rich and famous. The probability that a person picked at random from the population is either famous or rich but not both,is equal to

Similar Questions

If two events $E_1$ and $E_2$ are such that $P(E_1 \cup E_2) = \frac{5}{8}$, $P(\bar{E}_1) = \frac{3}{4}$, and $P(E_2) = \frac{1}{2}$, then $E_1$ and $E_2$ are:

If $P(A \cup B) = \frac{2}{3}$,$P(A \cap B) = \frac{1}{6}$,and $P(A) = \frac{1}{3}$,then which of the following is true?

Given two independent events $A$ and $B$ such that $P(A) = 0.3$ and $P(B) = 0.6$. Find $P(A \text{ or } B)$.

If $A$ and $B$ are independent events such that odds in favour of $A$ is $2:3$ and odds against $B$ is $4:5$,then $P(A \cap B)=$

It is $5:2$ against a husband who is $65$ years old living till he is $85$ and $4:3$ against his wife who is now $58$,living till she is $78$. If the probability that at least one of them will be alive for $20$ years is $k$,then the value of $49k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module