If A and B are two events such that P(A) = fr | vedclass

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

Question

(A) We are given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(A \cap B) = \frac{1}{5}$.First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) =

Vedclass · Accepted Answer

$\frac{37}{40}$

Vedclass · Answer

(A) We are given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(A \cap B) = \frac{1}{5}$.First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{5} = \frac{20 + 15 - 12}{60} = \frac{23}{60}$.We need to find $P\left( \frac{\overline{B}}{\overline{A}} ight)$. By the definition of conditional probability,$P\left( \frac{\overline{B}}{\overline{A}} ight) = \frac{P(\overline{B} \cap \overline{A})}{P(\overline{A})}$.Using De Morgan's Law,$\overline{B} \cap \overline{A} = \overline{A \cup B}$,so $P(\overline{B} \cap \overline{A}) = 1 - P(A \cup B) = 1 - \frac{23}{60} = \frac{37}{60}$.Also,$P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$.Therefore,$P\left( \frac{\overline{B}}{\overline{A}} ight) = \frac{37/60}{2/3} = \frac{37}{60} 	imes \frac{3}{2} = \frac{37}{40}$.

If $A$ and $B$ are two events such that $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$ and $P(A \cap B) = \frac{1}{5}$,then $P\left( \frac{\overline{B}}{\overline{A}} \right) = $

Similar Questions

$A$ bag contains $4$ white and $3$ red balls. Two draws of one ball each are made without replacement. Then the probability that both the balls are red is (in $/7$)

The probability that an event will fail to happen is $0.05$. The probability that the event will take place on $4$ consecutive occasions is

Six cards are drawn simultaneously from a pack of playing cards. What is the probability that $3$ will be red and $3$ will be black?

The event $A$ is independent of itself if and only if $P(A) = $

The probability of drawing a white ball from a bag containing $3$ black balls and $4$ white balls is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module