Eeny, meeny, miny, moe

The binary expansion of the given numbers has 10 digits so

The sample space is $512*\dbinom{10}{2}$

Consider any arbitrary number 1011110001=x assuming we are allowed to choose any 2 digits except the 1st digit we get $\dbinom{9}{2}$ different 10 bit numbers some of which are greater while some are smaller.

As our list contains all these 10 bit numbers we will have a case for each of the numbers obtained above by which we can go back to x.

eg. if we choose the last 2 bits in x we get the number 1011110010=y which is greater than x. now again if we chosse the last 2 digits for y we get x which is smaller than y.

So for every case in $512*\dbinom{9}{2}$ we have a complementary case in $512*\dbinom{9}{2}$

And if we choose the 1st digit as one of the two digits then the probablity of getting a bigger no is zero since the number obtained will be a 9 digit number.

so final probablity = $1/2*\frac{512*\dbinom{9}{2}}{512*\dbinom{10}{2}}=2/5$

Every element is a 10-digit binary number whose first digit is $1$ , and the set contains every 10-digit binary number. Additionally, $m > n$ iff the left digit swapped in $n$ is $0$ .

Instead of considering the questions by numbers, consider it by the digits. There are ${10 \choose 2} = 45$ ways to choose two digits. If one of the digits is the first, then the left digit is $1$ , so $n > m$ .

For the other $36$ combinations, there is an equal $\frac{1}{2}$ chance that the left most digit is a $0$ or $1$ .

Then the probability is $\frac{(36)(\frac{1}{2})}{45} = \frac {2}{5}$ .

Each number in the problem set is composed of 10 binary digits. Now divide the selection of the two digits into 2 cases:

1. The highest value digit is one of the ones chosen (with probability 9/45=1/5)
2. The highest value digit is not one of the ones chosen (with probability 36/45=4/5)

Now in case 1, the first digit is always a one, so it will be changed into a 0, and the statement will be false. So we are only concerned with case 2, and it's easy to see that the probability of the statement being true given case 2 is 1/2. Multiply that by the probability of case 2 which is 4/5, and you get 2/5. So the answer is 2+5= 7 .

(9C2/2)/(9C2+9) = a/b => 2/5 = a/b

512 in binary is 1 followed by 9 zeros. 1023 is 1 followed by 9 ones. When choosing a single number from 512 to 1023, the probability of the first binary digit being 1 is 1 and the probability of it being zero is 0. For the rest of the 9 binary digits, there is equal probability of it being a zero or a 1 (so probability of it being zero is 1/2 and probability of it being a 1 is 1/2). In the second stage, we choose two of the 10 digits. If we choose the first digit and one of the other 9 digits, then m > n with probability 0 (since that never happens). If both digits are chosen from the rest of the 9 digits (excluding the most significant digit), then the probability of m > n is 1/2. The probability of choosing both digits from the 9 least significant digits is 9C2/10C2 = 8/10. So the overall probability of m > n is 8/10 * 1/2 = 2/5.

The probability that $m > n$ is the probability that the more significant of the two digits chosen in $N$ is equal to $0$ . Let $X$ be the position of the more significant of the two digits (so that $X=1$ when the more significant digit chosen is the most significant digit in $N$ , and $X=9$ when the more significant digit chosen is the penultimate digit and the two digits chosen are the last two). Then $\mathbb{P}[X=j] \; = \; \tfrac{1}{45}(10-j) \qquad 1 \le j \le 10$ We have conditional probabilities $\mathbb{P}[m>n | X=j ] \; = \; \left\{ \begin{array}{ll} 0 & j = 1 \\ \tfrac12 & 2 \le j \le 9 \end{array} \right.$ and hence $\begin{array}{rcl} \mathbb{P}[m>n] & = & \tfrac12\sum_{j=2}^9 \tfrac{1} {45}(10-j) \; = \; \tfrac{1}{90}\sum_{j=1}^8 j \\ & = & \tfrac{1}{90}\times36 \; = \; \tfrac25 \end{array}$ so that $a+b=7$ .