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 odds against a certain event is $5 : 2$ and the odds in favour of another event is $6 : 5$. If both the events are independent, then the probability that at least one of the events will happen is
A die is thrown. Let $A$ be the event that the number obtained is greater than $3.$ Let $B$ be the event that the number obtained is less than $5.$ Then $P\left( {A \cup B} \right)$ is
The probability that a man will be alive in $20$ years is $\frac{3}{5}$ and the probability that his wife will be alive in $20$ years is $\frac{2}{3}$. Then the probability that at least one will be alive in $20$ years, is
Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is
If $A$ and $B$ are events such that $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ $P(\bar A) = 2/3,$ then $P(\bar A \cap B)$ is