Request a call back

doubt
Asked by armaanjain840 | 20 Dec, 2023, 05:59: PM

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

## Concept Videos

CBSE 12-science - Maths
Asked by armaanjain840 | 20 Dec, 2023, 05:59: PM
CBSE 12-science - Maths
Asked by sachinkumarmishra1 | 31 Jan, 2020, 05:17: PM
CBSE 12-science - Maths
Asked by sachinkumarmishra1 | 29 Jan, 2020, 02:59: PM
CBSE 12-science - Maths
Asked by sachinkumarmishra1 | 27 Jan, 2020, 12:27: PM
CBSE 12-science - Maths
Asked by sachinkumarmishra1 | 24 Jan, 2020, 11:10: AM
CBSE 12-science - Maths
Asked by sachinkumarmishra1 | 23 Jan, 2020, 12:00: PM