Question

In a hostel, 60% of the students read Hindi newspapers, 40% read English newspapers and 20% read both Hindi and English newspapers. A student is selected at random.
a.Find the probability that she reads neither Hindi nor English newspapers.
b.If she reads Hindi newspapers, find the probability that she reads English newspapers.
c.If she reads English newspapers, find the probability that she reads Hindi newspapers.
A.$0.56,0.78,0.76$
B.$0.20,0.33,0.50$
C.$0.65,0.45,0.34$
D.$0.56,0.56,0.65$

Answer 1

Hint: Try to name the group of students who read a specific newspaper, as naming will help you find the desired answer by putting it in a formula and then find the probability of occurring events.
Given: In the given question data of students reading a newspaper, it may be Hindi or English. It is given that 60% of the students read Hindi newspapers and 40% read English newspapers and 20%read both Hindi and English newspapers.

Complete step-by-step solution:
Let’s assume students who are reading Hindi newspapers are group$A$, and students who are reading English newspapers are group$B$.
It is given that the probability of selecting a student at random who is reading Hindi newspaper is $P\left( A \right) = \dfrac{{60}}{{100}}$or 0.60 and, the probability of selecting a student at random who is reading an English newspaper is $P\left( B \right) = \dfrac{{40}}{{100}}$or 0.40 , and probability of selecting a student at random who is reading both Hindi and English newspaper is $P\left( {A \cap B} \right) = \dfrac{{20}}{{100}}$or 0.20.
In this question we have to find the probability of the student selected reading neither Hindi nor English newspapers, for finding the probability we have to first find the probability of the student who read newspapers.
Therefore, a formula is used to find i.e.$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$, from this formula we will be able to find the probability of selecting a student who reads at least one newspaper.
Now, using the formula $P\left( {A \cup B} \right) = 0.60 + 0.40 - 0.20 = 0.80{\kern 1pt}$ (Equation$1$)
As we know, the probability of a student studying a newspaper and not studying a newspaper is 1.
So, Probability of neither studying a Hindi nor English newspaper$=$1- Probability of studying at least one newspaper (which we have found in equation$1$)
Therefore, the answer of a student not studying any newspaper would be$1 - 0.80 = 0.20$.

In this part, we have to find , if the selected student reads Hindi newspapers, then what is the probability that the selected student reads English newspapers. To find this we have to use a formula i.e.$P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$, in which the value of $P\left( {A \cap B} \right) = 0.20$and the value of$P\left( A \right) = 0.60$.
Therefore answer would be$P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = \dfrac{{0.20}}{{0.60}} = \dfrac{1}{3} = 0.33$.

In this part, we have to find the selected student who reads English newspapers, then what is the probability that the selected student reads Hindi newspapers. To find this we have to use a formula i.e.$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$, in which the value of $P\left( {A \cap B} \right) = 0.20$and the value of$P\left( B \right) = 0.40$.
Therefore, answer would be$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \dfrac{{0.20}}{{0.40}} = \dfrac{1}{2} = 0.50$.

So, after finding all the answers the correct option is B i.e. $0.20,0.33,0.50$.
Hence, the correct option is B.

Note: In this question we will use the formula of the probability when the events are not mutually exclusive, that is $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$.