Verified Solution • 2025–26 Session
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QJEEM20Mad58d: JEE Main 2020 Mathematics - Class 12 JEEM20Mad58d Solution: The probabilities of three events A, B and C are given by P(A)…
The probabilities of three events A, B and C are given by
P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
If P(A B) = 0.8, P(A C) = 0.3, P(A B C) = 0.2, P(B C) =
and P(A B C) = , where 0.85 0.95, then lies in the interval :
P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
If P(A B) = 0.8, P(A C) = 0.3, P(A B C) = 0.2, P(B C) =
and P(A B C) = , where 0.85 0.95, then lies in the interval :
Multiple Choice Options
- A [0.35, 0.36]
- B [0.20, 0.25]
- C [0.25, 0.35]
- D [0.36, 0.40]
Step-by-Step Verified Solution
P(A B) = P(A) + P(B) – P(A B)
0.8 = 0.6 + 0.4 – P(A B)
P(A B) = 0.2
P(A B C) = P(A) + P(B) + P(C) – P(A B) – P(B C) –P(C A) + P(A B C)
= 0.6 + 0.4 + 0.5 - 0.2 - - 0.3 + 0.2
+ = 1.2
= 1.2 -
Given, 0.85 0.95
0.85 1.2 - 0.95
0.25 0.35
0.8 = 0.6 + 0.4 – P(A B)
P(A B) = 0.2
P(A B C) = P(A) + P(B) + P(C) – P(A B) – P(B C) –P(C A) + P(A B C)
= 0.6 + 0.4 + 0.5 - 0.2 - - 0.3 + 0.2
+ = 1.2
= 1.2 -
Given, 0.85 0.95
0.85 1.2 - 0.95
0.25 0.35