OperatingSystems
October 21, 2023UGC NET CS 2013 Decpaper2
October 21, 2023DigitalLogicDesign
Question 2

Two numbers are chosen independently and uniformly at random from the set {1, 2, …, 13}. The probability (rounded off to 3 decimal places) that their 4bit (unsigned) binary representations have the same most significant bit is ______.
0.502


0.461


0.402


0.561

Question 2 Explanation:
Correct answer is 0.502
1 – 0001
2 – 0010
3 – 0011
4 – 0100
5 – 0101
6 – 0110
7 – 0111
8 – 1000
9 – 1001
10 – 1010
11 – 1011
12 – 1100
13 – 1101
The probability that their 4bit binary representations have the same most significant bit is
= P(MSB is 0) + P(MSB is 1)
= (7×7)/(13×13) + (6×6)/(13×13)
= (49+36)/169
= 85/169
= 0.502
1 – 0001
2 – 0010
3 – 0011
4 – 0100
5 – 0101
6 – 0110
7 – 0111
8 – 1000
9 – 1001
10 – 1010
11 – 1011
12 – 1100
13 – 1101
The probability that their 4bit binary representations have the same most significant bit is
= P(MSB is 0) + P(MSB is 1)
= (7×7)/(13×13) + (6×6)/(13×13)
= (49+36)/169
= 85/169
= 0.502
Correct Answer: A
Question 2 Explanation:
Correct answer is 0.502
1 – 0001
2 – 0010
3 – 0011
4 – 0100
5 – 0101
6 – 0110
7 – 0111
8 – 1000
9 – 1001
10 – 1010
11 – 1011
12 – 1100
13 – 1101
The probability that their 4bit binary representations have the same most significant bit is
= P(MSB is 0) + P(MSB is 1)
= (7×7)/(13×13) + (6×6)/(13×13)
= (49+36)/169
= 85/169
= 0.502
1 – 0001
2 – 0010
3 – 0011
4 – 0100
5 – 0101
6 – 0110
7 – 0111
8 – 1000
9 – 1001
10 – 1010
11 – 1011
12 – 1100
13 – 1101
The probability that their 4bit binary representations have the same most significant bit is
= P(MSB is 0) + P(MSB is 1)
= (7×7)/(13×13) + (6×6)/(13×13)
= (49+36)/169
= 85/169
= 0.502
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