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Question
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English news papers. A student is selected at random.
Find the probability that she reads neither Hindi nor English news papers.
Solution
Let H = event of reading Hindi newspaper,
E = event of reading an English newspaper
Then P(H) = `60/100 = 3/5`, P(E) = `40/100 = 2/5`
and P(H ∩ E) = `20/100 = 1/5`
Intended process= P(H' ∩ E') = P(H ∪ E)' = 1 − P(H ∪ E)
= 1 − [P(H) + P(E) − P(H ∩ E)]
= `1 - [3/5 + 2/5 - 1/5]`
= `1 - 4/5`
= `1/5`
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