$A$ and $B$ are events such that $P(A)=0.42$,  $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $A ).$

Fill in the blanks in following table :

Suppose that $A, B, C$ are events such that $P\,(A) = P\,(B) = P\,(C) = \frac{1}{4},\,P\,(AB) = P\,(CB) = 0,\,P\,(AC) = \frac{1}{8},$ then $P\,(A + B) = $

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 $)$

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

A coin is tossed twice. If events $A$ and $B$ are defined as :$A =$ head on first toss, $B = $ head on second toss. Then the probability of $A \cup B = $