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Two numbers are chosen independently and at random from set { 1,2.....13}. Find The probability that theor 4-bit unsigned binary representatives have the same most significant bit .

note: unsigned is way of representation of +ve numbers. I am giving the binary representation of unsigned 4 bit binary numbers from 0 -15.

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

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  • $\begingroup$ How did you get 6/13? @maveric $\endgroup$ Mar 3 '19 at 8:32
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Hint: how many have MSB of $0$? How many have MSB of $1$? How many ways to choose two that have the same MSB? The answer depends on whether the choice is with or without replacement, which you did not specify.

Added: there are seven numbers with MSB of $0$ and six with MSB of $1$. There are ${7 \choose 2}+{6 \choose 2}=21+15=36$ ways to choose two numbers with the same MSB. There are ${13 \choose 2}=78$ ways to choose two numbers overall, so the probability is $\frac {36}{78}=\frac {18}{39}$

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  • $\begingroup$ without replacement. i am getting 6/13 ans is 1/2 $\endgroup$
    – maveric
    Feb 9 '19 at 4:43
  • $\begingroup$ @maveric without replacement, use $n^2$ instead of $\binom n2$ in all the above calculations , since $n^2$ is the count of ways to make 2 independent selections each from $n$ options.. $\endgroup$ Jun 8 '19 at 3:15
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There are 7 numbers which are having the same MSB(0).

There are 6 numbers which are having the same MSB(1).

so the probability will be = $(\frac{7} {13} \times \frac{7} { 13}) + (\frac{6} { 13} \times \frac{6} {13})$ = $\frac{85} { 169}$ = $0.50295$

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