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

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

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

$=0.3-0.18$

$=0.12$

A fair coin and an unbiased die are tossed. Let $A$ be the event ' head appears on the coin' and $B$ be the event ' $3$ on the die'. Check whether $A$ and $B$ are independent events or not.

In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted neither $NCC$ nor $NSS$.

If the odds against an event be $2 : 3$, then the probability of its occurrence is

Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.7$, $P ( A \cap B )=0.6$

The probability that $A$ speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact

- [AIEEE 2004]