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Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

32. In a hostel 60% of the students read Hindi newspaper, 40% read English newspaper 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 newspaper, find the probability that she reads English newspaper.

(c) If she reads English newspapers, find the probability she reads Hindi newspaper.

3 Views | Posted 4 months ago

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Answered by

alok kumar singh | Contributor-Level 10

4 months ago

32. Let,

$P (H) =$ denote the student who read Hindi newspaper

$P (E) =$ denote the student who read English newspaper

Then, $P (H) = 60 % = \frac{6}{10} = \frac{3}{5}$

$P (E) = 40 % = \frac{4}{10} = \frac{2}{5}$

$i. P (H \cap E) = 20 % = \frac{2}{10} = \frac{1}{5}$

$\Rightarrow P (H \cup E)^{'} = 1 - [P (H) + P (E) - P (H \cap E)]$

$= 1 - [\frac{3}{5} + \frac{2}{5} - \frac{1}{5}]$

$= 1 - \frac{3 + 2 - 1}{5}$

$= 1 - \frac{4}{5} = \frac{1}{5}$

ii. Probability of randomly chosen student that reads English newspaper, if she reads Hindi newspaper, is given by

$P (E / H) = \frac{P (E \cap H)}{P (H)} = \frac{\frac{1}{5}}{\frac{3}{5}} = \frac{1}{3}$

iii. Probability that randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by

$P (H / E) = \frac{P (H \cap E)}{P (E)} = \frac{\frac{1}{5}}{\frac{2}{5}} = \frac{1}{2}$

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