If P(B) = frac{3}{4} , P(A cap B cap bar C) = frac{1}{3}{rm{ }} and P(bar A cap B cap bar C) = frac{

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14.Probability

medium

If $P(B) = \frac{3}{4}$, $P(A \cap B \cap \bar C) = \frac{1}{3}{\rm{ }}$ and $P(\bar A \cap B \cap \bar C) = \frac{1}{3},$ then $P(B \cap C)$ is

$\frac{1}{{12}}$

$\frac{1}{6}$

$\frac{1}{{15}}$

$\frac{1}{9}$

(IIT-2003)

Solution

(a) From Venn diagram, we can see that
$P(B \cap C)\, = P(B) – P(A \cap B \cap \bar C) – P(\bar A \cap B \cap \bar C)$
$ = \frac{3}{4} – \frac{1}{3} – \frac{1}{3} = \frac{1}{{12}}.$

Standard 11

Mathematics

If $A$ and $B$ are two events, then the probability of the event that at most one of $A, B$ occurs, is

normal

(IIT-2002)

The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B') = $

medium

Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$

easy

Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is found to be red, is

hard

(IIT-1988)

Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that First ball is black and second is red.