If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $
$P(\bar A)\,\,\,P(\bar B)$
$1 - P(A) - P(B)$
$P(A) + P(B) - P(A \cap B)$
$P(B) - P(A \cap B)$
(d) It is a fundamental concept.
Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is found to be red, is
If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P\,(F) < 1,$ then
$P(A \cup B) = P(A \cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is
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 $)$
Two dice are thrown. What is the probability that the sum of the numbers appearing on the two dice is $11$, if $5$ appears on the first
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