# Two numbers are chosen independently and at random.

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

• How did you get 6/13? @maveric Mar 3 '19 at 8:32

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 $$\frac {36}{78}=\frac {18}{39}$$
• @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.. Jun 8 '19 at 3:15
so the probability will be = $$(\frac{7} {13} \times \frac{7} { 13}) + (\frac{6} { 13} \times \frac{6} {13})$$ = $$\frac{85} { 169}$$ = $$0.50295$$