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CBSE Class 12-science Answered

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Asked by armaanjain840 | 20 Dec, 2023, 05:59: PM
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Since E1 , E2 , E3 are mutually exclusive events , then we have

p(E1) + p(E2) + p(E3) ≤ 1

26-3p ≤ 24   or  p ≥ (2/3)

p(E3) ≤ 1 , hence p ≤ 1

The condition p ≤ 1 is satisfied for other probabilities of mutually exclusive events so that p(E1) ≤ 1 and p(E2) ≤ 1 .

Hence minimum value of p is (2/3) and maximum vaue of p is 1

maximum of p = p1 = 1 

minimum of p = p2 = (2/3)

p1+ p2 = 1 + (2/3) = 5/3

Answered by Thiyagarajan K | 20 Dec, 2023, 09:28: PM

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CBSE 12-science - Maths
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Asked by armaanjain840 | 20 Dec, 2023, 05:59: PM
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