The probability that a student will pass the final examination in both English and Hindi is $0.5$ and the probability of passing neither is $0.1$. If the probability of passing the English examination is $0.75$, what is the probability of passing the Hindi examination?
Let $A$ and $B$ be the events of passing English and Hindi examination respectively.
Accordingly, $P ( A $ and $B)=0.5$, $P ($ not $A$ and $B )=0.1,$
i.e., $P \left( A^{\prime} \cap B ^{\prime}\right)=0.1$
$P ( A )=0.75$
Now, $P ( A \cap B ) ^{\prime}= P \left( A ^{\prime} \cap B ^{\prime}\right)$ [De Morgan's law]
$\therefore P(A \cap B)^{\prime}=P\left(A^{\prime} \cap B^{\prime}\right)=0.1$
$P ( A \cup B )=1- P ( A \cup B )^{\prime} =1-0.1=0.9$
We know that $P ( A$ or $ B )= P ( A )+ P ( B )- P ( A$ and $ B )$
$\therefore $ $0.9=0.75+ P ( B )-0.5$
$\Rightarrow P ( B )=0.9-0.75+0.5$
$\Rightarrow P(B)=0.65$
Thus, the probability of passing the Hindi examination is $0.65$.
The probability that a leap year selected at random contains either $53$ Sundays or $53 $ Mondays, is
For an event, odds against is $6 : 5$. The probability that event does not occur, is
If the odds against an event be $2 : 3$, then the probability of its occurrence is
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
Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that One of them is black and other is red.