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
$\frac{4}{9}$
$\frac{2}{3}$
$\frac{1}{3}$
$\frac{1}{2}$
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 $NSS$ but not $NCC$.
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') = $
Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that both balls are red.
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $A ).$