Let A and B be two events such that Poverline | vedclass

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

Question

(a) $P\left( {\overline {A \cup B} } \right) = \frac{1}{6};\,\,P\left( {A \cap B} \right) = \frac{1}{4}$,

$P\left( {\bar A} \right) = \frac{1}{4} \

Vedclass · Accepted Answer

Independent but not equally likely

Vedclass · Answer

(a) $P\left( {\overline {A \cup B} } ight) = \frac{1}{6};\,\,P\left( {A \cap B} ight) = \frac{1}{4}$,

$P\left( {\bar A} ight) = \frac{1}{4} \Rightarrow P\left( A ight) = \frac{3}{4}$,

$P\left( {\overline {A \cup B} } ight) = 1 - P\left( {A \cup B} ight) = 1 - P\left( A ight) - P\left( B ight) + P\left( {A \cap B} ight)$

$&rArr;$ $\frac{1}{6} = \frac{1}{4} - P\left( B ight) + \frac{1}{4}$ $&rArr;$ $P\left( B ight) = \frac{1}{3}$.

Since $P\left( {A \cap B} ight) = P\left( A ight)P\left( B ight)$ and $P\left( A ight) 
e P\left( B ight)$

$	herefore$ $ A$ and $B$ are independent but not equally likely.

$P(A)$	$P(B)$	$P(A \cap B)$	$P (A \cup B)$
$\frac {1}{3}$	$\frac {1}{5}$	$\frac {1}{15}$	........

Let $A$ and $B$ be two events such that $P\overline {(A \cup B)} = \frac{1}{6},P(A \cap B) = \frac{1}{4}$ and $P(\bar A) = \frac{1}{4},$ where $\bar A$ stands for complement of event $A$. Then events $A$ and $B$ are

Similar Questions

If $P(A) = 0.25,\,\,P(B) = 0.50$ and $P(A \cap B) = 0.14,$ then $P(A \cap \bar B)$ is equal to

Fill in the blanks in following table :

$P(A)$ $P(B)$ $P(A \cap B)$ $P (A \cup B)$

$\frac {1}{3}$ $\frac {1}{5}$ $\frac {1}{15}$ ........

$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .

One card is drawn randomly from a pack of $52$ cards, then the probability that it is a king or spade is

True statement $A$ and true statement $B$ are two independent events of an experiment.Let $P\left( A \right) = 0.3$ , $P\left( {A \vee B} \right) = 0.8$ then $P\left( {A \to B} \right)$ is (where $P(X)$ denotes probability that statement $X$ is true statement)