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Ex 13.2

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Ex 13.2, 11 (i)

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Ex 13.2, 16 Important You are here

Ex 13.2, 17 (MCQ)

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Chapter 13 Class 12 Probability

Serial order wise

Last updated at Aug. 16, 2023 by Teachoo

Ex 13.2, 16 In a hostel, 60% of the students read Hindi news paper, 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.Let H: Students reading Hindi newspaper & E: Students reading English newspaper Given, P(H) = 60% = 60/100 = 0.6 P(E) = 40% = 40/100 = 0.4 P(H "∩" E) = 20% = 20/100 = 0.2 We need to find the Probability that she reads neither Hindi nor English newspapers. i.e. P(H’ "∩" E’) Now, P(H’ "∩" E’) = 1 – Probability that she reads both the newspapers. = 1 – P(H ∪ E) = 1 − [P(H) + P(E) – P(H ∩ E)] = 1 − [0.6 + 0.4 – 0.2] = 1 – 0.8 = 0.2 = 𝟏/𝟓 Ex 13.2, 16 In a hostel, 60% of the students read Hindi news paper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (b) If she reads Hindi news paper, find the Probability that she reads English newspaper.We need to find the Probability that she reads English newspaper, given she reads Hindi newspaper i.e. we need to find P(E|H) Now, P(E|H) = "P(E ∩ H)" /"P(H)" = (0. 2)/(0. 6) = 2/6 = 𝟏/𝟑 Ex 13.2, 16 In a hostel, 60% of the students read Hindi news paper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (c) If she reads English news paper, find the Probability that she reads Hindi newspaper.We need to find the Probability that she reads Hindi newspaper, given she reads English i.e. P(H|E) Now, P(H|E) = "P(H ∩ E)" /"P(E)" = (0. 2)/(0. 4) = 2/4 = 𝟏/𝟐