Question
Consider a $4$ bit binary representation of decimal number ranging $1$ to $13$.
What is the probability that 2 numbers selected have same msb?
My Approach
Converting the decimal number to $4$ bit representation from $1$ to $13$,we have
$0001$,$0010$,$0011$,$0100$,$0101$,$0110$,$0111$
$1000$,$1001$,$1010$,$1011$,$1100$,$1101$
If we select number whose msb is $0$,we have $6$ options and if we select number
whose msb is $1$,we have $5$ options
so reqd probability$=\frac{6 \times 5}{\binom{13}{2}}=\frac{5}{13}$
Is it correct?
Please help
Best Answer
There are $\binom72$ ways to choose $0$ as the most significant bit for a pair and $\binom62$ ways to choose $1$ as the most significant bit for a pair.
Hence it should be $$\frac{\binom72 + \binom62}{\binom{13}2}$$