Zoran Skoda expectation of number of valid tickets

Suppose that there are kk valid and ll invalid tickets in a pool and we choose nn tickets from the pool at random. What is the expectation of the number of tickets ?

The probability that a randomly chosen ticket is valid is kk+l\frac{k}{k+l}, so that the expectation for the number vv of valid tickets in the set of nn is nkk+ln\frac{k}{k+l}. Indeed, this is the expectation of the sum of validity points for nn tickets.

We can directly find the expectation by the combinatorics. Indeed, the probability to have ii valid tickets in the set of nn is

P({v=i})=(ki)(lni)(k+ln),P(\{v=i\}) = \frac{\binom{k}{i}\binom{l}{n-i}}{\binom{k+l}{n}},

so that the expectation is

E[v]= i=0 ni(ki)(lni)(k+ln)=nkk+l E[v] = \frac{\sum_{i=0}^n i\binom{k}{i}\binom{l}{n-i}} {\binom{k+l}{n}} = n\frac{k}{k+l}

thanks to the identity

i=0 ni(ki)(lni)=nkk+l(k+ln) \sum_{i=0}^n i\binom{k}{i}\binom{l}{n-i} = \frac{n k}{k+l}\binom{k+l}{n}

For example, for k=8k=8, l=13l=13 the sum

i=0 6(8i)(136n) \sum_{i = 0}^6 \binom{8}{i}\binom{13}{6-n}

can be evaluated as

628+55613+47078+356286+228715+181287+01716 6\cdot 28+5\cdot 56\cdot 13+4\cdot 70\cdot 78+3\cdot 56\cdot 286+2\cdot 28\cdot 715+1\cdot 8\cdot 1287+0\cdot 1716

what is equal to

124032=16754264=6821(216) 124032 = \frac{16}{7} 54264 = \frac{6\cdot 8}{21}\binom{21}{6}

Thus, E[v]=16/7E[v] = 16/7. Similarly,

i=0 4i(8i)(244i)=(324) \sum_{i=0}^4 i \binom{8}{i}\binom{24}{4-i} = \binom{32}{4}
i=0 4i(4i)(284i)=12(324) \sum_{i=0}^4 i \binom{4}{i}\binom{28}{4-i} = \frac{1}{2}\binom{32}{4}

Created on November 20, 2020 at 00:12:59. See the history of this page for a list of all contributions to it.