Given two independent events $A$ and $B$ such that $P(A) $ $=0.3, \,P(B)=0.6$ Find $P(A$ and $B)$.
It is given that $P(A)=0.3,\,P(B)=0.6$..
Also, $A$ and $B$ are independent events.
$\mathrm{P}(\mathrm{A}$ and $\mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
$\Rightarrow $ $ \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.3 \times 0.6=0.18$
Let $A$ and $B$ be two events such that $P\,(A) = 0.3$ and $P\,(A \cup B) = 0.8$. If $A$ and $B$ are independent events, then $P(B) = $
If $A$ and $B$ are two events such that $P(A) = 0.4$ , $P\,(A + B) = 0.7$ and $P\,(AB) = 0.2,$ then $P\,(B) = $
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to
An electronic assembly consists of two subsystems, say, $A$ and $B$. From previous testing procedures, the following probabilities are assumed to be known :
$\mathrm{P}$ $( A$ fails $)=0.2$
$P(B$ fails alone $)=0.15$
$P(A$ and $ B $ fail $)=0.15$
Evaluate the following probabilities $\mathrm{P}(\mathrm{A}$ fails alone $)$
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 $NSS$.