 QUESTION

# The probability that a student will pass the 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?

Hint: Probability of event E = $\dfrac{n(E)}{n(S)}=\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Use the fact that $n\left( A\bigcup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\bigcap B \right)$. Use de morgan's law and hence the probability of passing either of the exams. Use this value to find the probability of passing the Hindi exam.

Let A be the event: Student passes English.
Let B be the event: Student passes Hindi.
We know that $n\left( A\bigcup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\bigcap B \right)$
Dividing both sides by n(S), we get
\begin{align} & \dfrac{n\left( A\bigcup B \right)}{n\left( S \right)}=\dfrac{n\left( A \right)}{n\left( S \right)}+\dfrac{n\left( B \right)}{n\left( S \right)}-\dfrac{n\left( A\bigcap B \right)}{n\left( S \right)} \\ & \Rightarrow P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right) \\ \end{align}
Now we have P(A) = 0.75, $P\left( A\bigcap B \right)=0.5$ and $P\left( A'\bigcap B' \right)=0.1$.
Using $P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$, we get
\begin{align} & P\left( A\bigcup B \right)=0.75+P\left( B \right)-0.5 \\ & \Rightarrow P\left( A\bigcup B \right)=0.25+P\left( B \right)\text{ (i)} \\ \end{align}
Also, we know from de morgan's law $\left( A\bigcup B \right)'=A'\bigcap B'$
Hence, we have $P\left( \left( A\bigcup B \right)' \right)=0.1$
We know that $P(E')=1-P(E)$.
Using the above formula, we get
$P\left( A\bigcup B \right)=1-P\left( \left( A\bigcup B \right)' \right)=1-0.1=0.9$
Hence from equation (i), we have
\begin{align} & 0.9=P\left( B \right)+0.25 \\ & \Rightarrow P\left( B \right)=0.9-0.25=0.65 \\ \end{align}
Hence the probability of passing the Hindi examination is 0.65.
Hence the probability that the chosen wristwatch is defective = 0.1

Note: The probability of an event always lies between 0 and 1
 The sum of probabilities of an event E and its complement E' = 1
i.e. $P(E)+P(E')=1$
Hence, we have $P(E')=1-P(E)$. This formula is applied when it is easier to calculate P(E') instead of (E).
This result can be proved by the fact that n(E)+n(E') =n(S)
Dividing both sides by n(S) gives the result.
 De morgan's laws for Union and Intersection of sets are
$\left( A\bigcup B \right)'=A'\bigcap B'$ and $\left( A\bigcap B \right)'=A'\bigcup B'$.