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JEE Main 2017

JEE Main 2017 Mathematics question on Probability

For three events AA, BB and CC, if PP(exactly one of AA or BB occurs) =P= P(exactly one of BB or CC occurs) =P= P(exactly one of CC or AA occurs) =14= \frac{1}{4} and PP(all the three events occur simultaneously) =116= \frac{1}{16}, then the probability that atleast one of the events occurs, is
  1. A.
    732\frac{7}{32}
  2. B.
    716\frac{7}{16}
  3. C.
    764\frac{7}{64}
  4. D.
    316\frac{3}{16}
Show answer and solution

Correct option: B

Solution

The key identity: P(exactly one of A or B) = P(A) + P(B) - 2P(A∩B) = 1/4.

Similarly for the other pairs. Adding all three equations:
2[P(A) + P(B) + P(C)] - 2[P(A∩B) + P(B∩C) + P(C∩A)] = 3/4

So P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A) = 3/8.

Now use inclusion-exclusion for P(at least one):
P(A∪B∪C) = P(A)+P(B)+P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C)
= 3/8 + 1/16
= 6/16 + 1/16 = 7/16.

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