Probability Ques 24
Let $S=(1,2, \ldots, 20)$. $A$ subset $B$ of $S$ is said to be “nice”, if the sum of the elements of $B$ is $203$. Then, the probability that a randomly chosen subset of $S$ is “nice”, is
(a) $\frac{6}{2^{20}}$
(b) $\frac{4}{2^{20}}$
(c) $\frac{7}{2^{20}}$
(d) $\frac{5}{2^{20}}$
(2019 Main, 11 Jan II)
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Answer:
Correct Answer: 24.(d)
Solution:
Formula:
- Number of subset of $S=2^{20}$
Sum of elements in $S$ is $1+2+\ldots .+20=\frac{20(21)}{2}=210$
$ [\because 1+2+\ldots \ldots+n=\frac{n(n+1)}{2}] $
Clearly, the sum of elements of a subset would be $203$, if we consider it as follows
$S-(7), S-(1,6) S-(2,5), S-(3,4)$, $S-(1,2,4)$
$\therefore$ Number of favourables cases $=5$
Hence, required probability $=\frac{5}{2^{20}}$
BETA
