Let S = {1, 2, … , 20). A and B subset 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

# Let S = {1, 2, ... , 20). A and B subset 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

1. A

$5/{2}^{20}$

2. B

$7/{2}^{20}$

3. C

$4/{2}^{20}$

4. D

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### Solution:

$S=\left\{1,2,\dots ,20\right\}$

Total no. of subsets =${2}^{20}$

We have, $1+2+\dots +20=\frac{20\left(21\right)}{2}=210$

Sum of elements will be 203 if we would not choose 7 or elements which are giving sum 7.

Elements giving sum 7 are (1, 6), (2, 5), (3, 4), (1, 2, 4) Thus, a subset will be 'nice' if we didn't choose these 5 outcomes.

$\therefore$  Required probability (subset choosen is 'nice') $=\frac{5}{{2}^{20}}$

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