Example — At least one jack

Example — 32 cards: the complementary rule

From a deck of $32$ cards, $5$ cards are drawn. What is the probability of drawing at least one jack?

Strategy. The complement of “at least one jack” is “no jack”. It is convenient to count the ways with no jacks and subtract them from the total number of ways:

$\#\,\text{modi totali} = \binom{32}{5}, \qquad \#\,\text{modi senza fanti} = \binom{28}{5}.$

$P(\text{almeno un fante}) = 1 - \frac{\binom{28}{5}}{\binom{32}{5}}.$

The “at least one” case is almost always computed via the complement: direct enumeration (exactly $1$ jack $+$ exactly $2$ $+\ldots+$ exactly $4$) would instead require $4$ terms.

Links

Topics: Probability
Concepts: Simple combinations · Complementary event · Probability
Methods: Simple combinations · Complementary prob
Skills: Combinatorial calculus · Probability calculation