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A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is almost 0.3. Is it possible that the probability of B getting selected is 0.9?

Jitender Pal
7 years ago

Hello Student,
Let A denote the event that the candidate A is selected and B the event that B is selected. It is given that
P (A) = 0.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
P (A ∩ B) ≤ 0.3 . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now, P (A) + P (B) - P (A ∩ B) = P (A ∪ B) ≤ 1
Or 0.5 + P (B) - P (A ∩ B) ≤ 1 [Using (1)]
Or P (B) ≤ 0.5 + P (A ∩ B) ≤ 0.5 + 0.3 [Using (2)]
Or P (B) ≤ 0.8 ∴ P (B) can not be 0.9

Thanks
Jitender Pal

Rahul Jain
13 Points
5 years ago
Please confirm, what’s wrong in this?
P(A and B) = P(A) × P(B)
0.3 = 0.5 × P(B)
P(B) = 0.3/0.5 = 0.6

Rahul Jain
Rishi Sharma
one year ago
Dear Student,

Let A denote the event that the candidate A is selected and B the event that B is selected.
It is given that P (A) = 0.5 . . . (1)
P (A ∩ B) ≤ 0.3 . . . . . (2)
Now, P (A) + P (B) - P (A ∩ B) = P (A ∪ B) ≤ 1
0.5 + P (B) - P (A ∩ B) ≤ 1 [Using (1)]
P (B) ≤ 0.5 + P (A ∩ B) ≤ 0.5 + 0.3 [Using (2)]
P (B) ≤ 0.8
∴ P (B) can not be 0.9

Thanks and Regards