1991 Paper 3 Q15

Year: 1991
Paper: 3
Question Number: 15

Course: LFM Stats And Pure
Section: Tree Diagrams

Difficulty: 1700.0 Banger: 1485.9

probability combinatorics cards without replacement hypergeometric counting arguments pairs tree diagram

Problem

A pack of \(2n\) (where \(n\geqslant4\)) cards consists of two each of \(n\) different sorts. If four cards are drawn from the pack without replacement show that the probability that no pairs of identical cards have been drawn is \[ \frac{4(n-2)(n-3)}{(2n-1)(2n-3)}. \] Find the probability that exactly one pair of identical cards is included in the four. If \(k\) cards are drawn without replacement and \(2 < k < 2n,\) find an expression for the probability that there are exactly \(r\) pairs of identical cards included when \(r < \frac{1}{2}k.\) For even values of \(k\) show that the probability that the drawn cards consist of \(\frac{1}{2}k\) pairs is \[ \frac{1\times3\times5\times\cdots\times(k-1)}{(2n-1)(2n-3)\cdots(2n-k+1)}. \]

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1485.9

Banger Comparisons: 3

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Problem source

A pack of $2n$ (where $n\geqslant4$) cards consists of two each
of $n$ different sorts. If four cards are drawn from the pack without
replacement show that the probability that no pairs of identical cards
have been drawn is 
\[
\frac{4(n-2)(n-3)}{(2n-1)(2n-3)}.
\]
Find the probability that exactly one pair of identical cards is included
in the four. 

If $k$ cards are drawn without replacement and $2 < k < 2n,$ find an
expression for the probability that there are exactly $r$ pairs of
identical cards included when $r < \frac{1}{2}k.$ 

For even values of $k$ show that the probability that the drawn cards
consist of $\frac{1}{2}k$ pairs is 
\[
\frac{1\times3\times5\times\cdots\times(k-1)}{(2n-1)(2n-3)\cdots(2n-k+1)}.
\]

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