Question

1) In a hostel, $60\%$ of the students read Hindi newspaper, $40\%$ read English news paper and $20\%$ read both Hindi and English newspapers. A student is selected at random.

(i) Find the probability that she reads neither Hindi nor English news papers.

2) In a hostel, $60\%$ of the students read Hindi newspaper, $40\%$ read English news paper and $20\%$ read both Hindi and English newspapers. A student is selected at random.

(ii) If she reads Hindi newspaper, find the probability that she reads English newspaper.

3) In a hostel, $60\%$ of the students read Hindi newspaper, $40\%$ read English news paper and $20\%$ read both Hindi and English newspapers. A student is selected at random.

(iii) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Answer 1

1) Let $E$ denote the students who read Hindi newspaper and $H$ denotes the students who read English newspaper.
From Question,
$P\left(E\right)=\frac{60}{100},P\left(F\right)=\frac{40}{100}$

$P(E\cap F)$

$=P\left(\text{Students who read both Hindi and English newspaper}\right)$

$P(E\cap F)=\frac{20}{100}$

We know that,

$P(E^{\prime }\cap F^{\prime })=1-\left[\begin{array}{c}P\left(E\right)+P\left(F\right)\\ -P(E\cap F)\end{array}\right]...\left(1\right)$

Substituting the value of $P\left(E\right),P\left(F\right)\&P(E\cap F)\text{in}\left(1\right),$

$P(E^{\prime }\cap F^{\prime })=1-[\frac{60}{100}+\frac{40}{100}-\frac{20}{100}]$

$P(E^{\prime }\cap F^{\prime })=1-\frac{4}{5}$

$\therefore P(E^{\prime }\cap F^{\prime })=\frac{1}{5}$

Therefore, the probability that she reads neither Hindi or English newspaper $=\frac{1}{5}$

2) Let $E$ denotes the students who read Hindi newspaper and $F$ denotes the students who read English newspaper.
From Question,

$P\left(E\right)=\frac{60}{100},P\left(F\right)=\frac{40}{100}$

$P(E\cap F)$

$=P\left(\text{Students who read both Hindi and English newspaper}\right)$

$P(E\cap F)=\frac{20}{100}$

Now, $P\left(F|E\right)$ = The prpbability that she reads English newspaper if she read Hindi newspaper

We know that,

$P\left(F|E\right)=\frac{P(E\cap F)}{P\left(E\right)}...\left(1\right)$

Substituting the value of $P\left(E\right)\&P(E\cap F)\text{in}\left(1\right),$

$P\left(E|F\right)=\left(\frac{\frac{20}{100}}{\frac{60}{100}}\right)$

$\therefore P\left(E|F\right)=\frac{1}{3}$

Therefore, the probability that she reads English newspaper if she read Hindi newspaper $=\frac{1}{3}$

3) Let $E$ denotes the students who read Hindi newspaper and $F$ denotes the students who read English newspaper.

From Question,

$P\left(E\right)=\frac{60}{100},P\left(F\right)=\frac{40}{100}$

$P(E\cap F)$

$=P\left(\text{Students who read both Hindi and English newspaper}\right)$

$P(E\cap F)=\frac{20}{100}$

Now, $P\left(E|F\right)=$ The probability that she reads Hindi newspaper if she read English newspaper

We know that

$P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}...\left(1\right)$

Substituting the value of $P\left(F\right)\&P(E\cap F)\text{in}\left(1\right),$

$P\left(E|F\right)=\left(\frac{\frac{20}{100}}{\frac{40}{100}}\right)$

$\therefore P\left(E|F\right)=\frac{1}{2}$

Therefore, the probability that she reads Hindi newspaper is she reads English newspaper $=\frac{1}{2}$