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 $)$
Let the event in which $A$ fails and $B $ fails be denote by $E_{A}$ and $E_{B}$.
$P\left(E_{A}\right)=0.2$
$\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \text { and } \mathrm{E}_{\mathrm{B}}\right)=0.15$
$\mathrm{P}(\mathrm{B} \text { fails alone })=\mathrm{P}\left(\mathrm{E}_{\mathrm{B}}\right)-\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \text { and } \mathrm{E}_{\mathrm{B}}\right)$
$\therefore $ $ 0.15=P\left(E_{B}\right)-0.15$
$\therefore $ $ \mathrm{P}\left(\mathrm{E}_{\mathrm{B}}\right)=0.3$
$\mathrm{P}$ $(A$ fails alone $)$ $=\mathrm{P}\left(\mathrm{E}_{\mathrm{A}}\right)-\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \text { and } \mathrm{E}_{\mathrm{B}}\right)$
$=0.2-0.15$
$=0.05$
If $P\,(A) = \frac{1}{4},\,\,P\,(B) = \frac{5}{8}$ and $P\,(A \cup B) = \frac{3}{4},$ then $P\,(A \cap B) = $
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 = $
Let $S$ be a set containing n elements and we select $2$ subsets $A$ and $B$ of $S$ at random then the probability that $A \cup B = S$ and $A \cap B = \phi $ is
One card is drawn at random from a well shuffled deck of $52$ cards. In which of the following cases are the events $\mathrm{E}$ and $\mathrm{F}$ independent ?
$E:$ 'the card drawn is a spade'
$F:$ 'the card drawn is an ace'
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