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

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.

### 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') $=\frac{5}{{2}^{20}}$

## Related content

 How to Score 100 in Class 6 Maths using NCERT Solutions TS EAMCET Previous Year Question Papers CBSE Class 8 English Syllabus Academic Year 2023-2024 CBSE Class 7 English Syllabus Academic Year 2023-2024 CBSE Worksheets for Class 7 with Answers COMEDK UGET Mock Test 2024 (Available) – Free Mock Test Series Indian tribes Maurya Empire CBSE Class 10 Science Important Topics – You Should Not Miss in Board Exam 2024 CBSE Class 6 Social Science: Important Tips and Topics

Talk to our academic expert!

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.