If $P(A) = 0.25,\,\,P(B) = 0.50$ and $P(A \cap B) = 0.14,$ then $P(A \cap \bar B)$ is equal to
$0.61$
$0.39$
$0.48$
None of these
(d) $P(A \cap \bar B) = P(A) – P(A \cap B) = 0.25 – 0.14 = 0.11$.
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
If $P(A) = 2/3$, $P(B) = 1/2$ and ${\rm{ }}P(A \cup B) = 5/6$ then events $A$ and $B$ are
In a class of $125$ students $70$ passed in Mathematics, $55$ in Statistics and $30$ in both. The probability that a student selected at random from the class has passed in only one subject is
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .
The probability that a leap year selected at random contains either $53$ Sundays or $53 $ Mondays, is
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