Let A and B be events with P(A)=frac{1}{3}, P | vedclass

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

Question

(C) Given,$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$.First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A

Vedclass · Accepted Answer

$P(A^C \cap B)=\frac{1}{3}$

Vedclass · Answer

(C) Given,$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$.First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{2} = \frac{4+3-6}{12} = \frac{1}{12}$.Check for independence: $P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$. Since $P(A \cap B) = P(A) \cdot P(B)$,$A$ and $B$ are independent. (Option $A$ is correct).Calculate $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/12}{1/4} = \frac{1}{3}$. (Option $B$ is correct).Calculate $P(A^C \cap B) = P(B) - P(A \cap B) = \frac{1}{4} - \frac{1}{12} = \frac{3-1}{12} = \frac{2}{12} = \frac{1}{6}$. (Option $C$ is incorrect,as it states $1/3$).Calculate $P(A \cap B^C) = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{12} = \frac{4-1}{12} = \frac{3}{12} = \frac{1}{4}$. (Option $D$ is correct).Thus,the incorrect statement is $P(A^C \cap B) = \frac{1}{3}$.

Let $A$ and $B$ be events with $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}$ and $P(A \cup B)=\frac{1}{2}$. Then which of the following statements is incorrect?

Similar Questions

Two dice are thrown and the sum of the numbers appeared on the dice is noted. If $A$ is the event of getting a prime number as their sum and $B$ is the event of getting a number greater than $8$ as their sum,then $P(A \cap \overline{B})=$

The probability that Krishna will be alive in $10$ years is $7/15$ and the probability that Hari will be alive in $10$ years is $7/10$. What is the probability that both will die within the next $10$ years (in $/150$)?

$A$ card is selected from a pack of $52$ cards. Calculate the probability that the card is an ace of spades.

The probability that an ordinary or a non-leap year has $53$ Sundays is

$4$-digit numbers are formed using the digits $4, 5, 6, 7, 8, 9$ allowing repetition of the given digits. If a number is chosen at random from those numbers thus formed, then the probability that it is exactly divisible by $3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module