The question is to evaluate $$\frac{\sum_{i=0}^{100} \binom{k}{i} \binom{M-k}{100-i} \frac{k-i}{M-100}}{\binom{M}{100}}$$
I found this question in Test of Maths at 10+2level. I tried writing down a few terms. I splitted the summation which involved only the $\sum_{i=0}^{100} \binom{k}{i} \binom{M-k}{100-i}$ which is the coefficient of $x^{100}$ in expansion of $(1+x)^M$ but i couldn't proceed after this. Any ideas? Thanks.
I assume $M\geq k + 100$ and $k\geq 101$, so that the various binomials are well defined. First of all let us observe that $$\binom{k}{i} (k-i) = k \binom{k-1}{i}.$$
Then the numerator becomes: $$\frac{k}{M-100}\sum_{i=0}^{100}\binom{k-1}{i} \binom{M-k}{100-i}.$$
Applying the Chu-Vandermonde identity we can evaluate the sum: $$\sum_{i=0}^{100}\binom{k-1}{i} \binom{M-k}{100-i} = \binom{M-1}{100}.$$
Therefore, the whole fraction that needs to be evaluated is now $$\frac{\sum_{i=0}^{100} \binom{k}{i} \binom{M-k}{100-i} \frac{k-i}{M-100}}{\binom{M}{100}} = \frac{\frac{k}{M-100}\binom{M-1}{100}}{\binom{M}{100}} = \frac{k}{M}.$$