Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

It is given that $\mathrm{P}(\mathrm{A})=0.3, \,\mathrm{P}(\mathrm{B})=0.6$

Also, $A$ and $B$ are independent events.

$\mathrm{P}(\mathrm{A}$ and not $\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})^{\prime}$

$=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$=0.3-0.18$

$=0.12$

Similar Questions

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

If $P(A) = \frac{1}{2},\,\,P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{7}{{12}},$ then the value of $P\,(A' \cap B')$ is

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

Two dice are thrown simultaneously. The probability that sum is odd or less than $7$ or both, is

If $A$ and $B$ are events such that $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ $P(\bar A) = 2/3,$ then $P(\bar A \cap B)$ is

  • [AIEEE 2002]