For three events $\mathrm{A}, \mathrm{B}$ and C , P(Exactly one of A or B occurs) $=\mathrm{P}($ Exactly one of B or C occurs $)$ $=\mathrm{P}($ Exactly one of C or A occurs $)=\frac{1}{4}$ and P(All the three events occur simultaneously) $=\frac{1}{16}$. Then the probability that at least one of the events occurs, is :

JEE Main · 2017 · 02 Apr 2017 · Offline · Probability

For three events

A, B

and C ,

P(Exactly one of A or B occurs)

= P (

Exactly one of B or C occurs

)

= P (

Exactly one of C or A occurs

) = \frac{1}{4}

and
P(All the three events occur simultaneously)

= \frac{1}{16}

.

Then the probability that at least one of the events occurs, is :

Answer: (C)

Write

P (A) = a

P (B) = b

P (C) = c

, with pairwise intersections

x, y, z

and triple

\frac{1}{16}

. "Exactly one of

A

B

" means

a + b - 2 x = \frac{1}{4}

, and likewise for the other two pairs. Adding the three relations,

2 (a + b + c) - 2 (x + y + z) = \frac{3}{4}

, so

(a + b + c) - (x + y + z) = \frac{3}{8}

. By inclusion-exclusion

P (A \cup B \cup C) = (a + b + c) - (x + y + z) + \frac{1}{16} = \frac{3}{8} + \frac{1}{16} = \frac{7}{16}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option