For three events A, B, and C of a sample spac | vedclass

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

Question

(C) Let $P(A) = x$, $P(B) = y$, $P(C) = z$, $P(A \cap B) = p$, $P(B \cap C) = q$, $P(C \cap A) = r$, and $P(A \cap B \cap C) = k = \frac{1}{16}$.Given

Vedclass · Accepted Answer

$\frac{7}{16}$

Vedclass · Answer

(C) Let $P(A) = x$, $P(B) = y$, $P(C) = z$, $P(A \cap B) = p$, $P(B \cap C) = q$, $P(C \cap A) = r$, and $P(A \cap B \cap C) = k = \frac{1}{16}$.Given $P(	ext{exactly one of } A 	ext{ or } B) = P(A) + P(B) - 2P(A \cap B) = x + y - 2p = \frac{1}{4}$.Similarly, $y + z - 2q = \frac{1}{4}$ and $z + x - 2r = \frac{1}{4}$.Adding these three equations: $2(x + y + z) - 2(p + q + r) = \frac{3}{4} \implies x + y + z - (p + q + r) = \frac{3}{8}$.The probability that at least one event occurs is $P(A \cup B \cup C) = (x + y + z) - (p + q + r) + k$.Substituting the values: $P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16} = \frac{6+1}{16} = \frac{7}{16}$.

Similar Questions

If $A$ and $B$ are two independent events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$,then which of the following is true?

Two persons $A$ and $B$ take turns in throwing a pair of dice. The first person to throw a sum of $9$ with both dice will win the game. If $A$ throws first,then the probability that $B$ wins the game is:

In a throw of three dice,the probability that at least one die shows up $1$,is

The probability that the product of the outcomes when three dice are rolled simultaneously is divisible by $4$ is equal to

Bag $A$ contains $3$ white and $4$ red balls,bag $B$ contains $4$ white and $5$ red balls,and bag $C$ contains $5$ white and $6$ red balls. If one ball is drawn at random from each of these three bags,then the probability of getting one white and two red balls is

Vedclass Test Series

Exam Paper Generator

Online Exam Module