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) =1/4 and P(All the three events occur simultaneously) =1/16. Then the probability that at least one of the events occurs, is:

Correct answer: (B) 7/16. a=1/16 c+d+e+f=1/4 ...i b+d+e+g=1/4 .....ii b+f+c+g=1/4 ....iii Adding i, ii and iii ⇒ 2b+c+d+e+f+g=3/4 ⇒ b+c+d+e+f+g=3/8 Now ⇒ PA∪B∪C=a+b+c+d+e+f+g ⇒ PA∪B∪C=1/16+3/8 ⇒ PA∪B∪C=7/16

Q34 / 90 · JEE Main 2017 (02 Apr) · Mathematics

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:

A
$\frac{7}{32}$
Correct answer
B
$\frac{7}{16}$
C
$\frac{7}{64}$
D
$\frac{3}{16}$

Solution

$a = \frac{1}{16}$

$c + d + e + f = \frac{1}{4} . . . (i)$

$b + d + e + g = \frac{1}{4} . . . . . (i i)$

$b + f + c + g = \frac{1}{4} . . . . (i i i)$

Adding $(i), (i i)$ and $(i i i)$

$\Rightarrow 2 (b + c + d + e + f + g) = \frac{3}{4}$

$\Rightarrow b + c + d + e + f + g = \frac{3}{8}$

Now $\Rightarrow P (A \cup B \cup C) = a + (b + c + d + e + f + g)$

$\Rightarrow P (A \cup B \cup C) = \frac{1}{16} + \frac{3}{8}$

$\Rightarrow P (A \cup B \cup C) = \frac{7}{16}$

Hence, the correct answer is option: 'B'.

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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) =14 and P(All the three events occur simultaneously) =116.Then the probability that at least one of the events occurs, is:

Solution

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