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GATE CSE 2019 | Question: 22

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Two numbers are chosen independently and uniformly at random from the set $\{1,2,\ldots,13\}.$
The probability (rounded off to $3$ decimal places) that their $4\text{-bit}$ (unsigned) binary representations have the same most significant bit is ___________.
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7 votes
7 votes

If we do $$\frac{{^7C_2} + {^6C_2}}{{^{13}C_2}}$$

it won't be correct as by doing this we ensure that in choosing 2 elements, they both will necessarily be distinct.

But to be independent, we should be able to select an item n number of times in n trials. Because choosing the same item before shouldn't matter in the present, as the trials are independent. We need to allow repetitions.

 

So,

$$\frac{7}{13}*\frac{7}{13}+\frac{6}{13}*\frac{6}{13}$$

will be correct.

 

0.503 is the answer.

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5 votes
5 votes

Ans will be 0.462 only!!

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3 votes
3 votes
1 to 7 have the most significant digit as 0. Rest 6 have 1 as MSB.

 

So, $((7*7) + (6 * 6) )/ (13 * 13)$

which is 0.5029
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1 votes
1 votes
As there are only 6 numbers in set A out of 13 numbers whose MSB will be 1 ( from 8-13) so the probility is 6/13 i.e. 0.4615
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