Question

A deck of cards contains 30 cards with labels 1, 2, . . . , 30. Suppose that somebody is randomly dealt a set of 7 cards of these cards (numbered with seven distinct numbers). a) Find the probability that 3 of the cards contain odd numbers and 4 contain even numbers. b) Find the probability each of the numbers on the seven cards ends with a different digit. (For example, the cards could be 3, 5, 14, 16, 22, 29, 30.

Answer 1

$\displaystyle |\Omega|=\binom{30}{7}=\dfrac{30!}{7!23!}=\dfrac{24\cdot25\cdot\ldots\cdot30}{2\cdot3\cdot\ldots\cdot 7}=2035800$

a)

$\displaystyle\\|A|=\binom{15}{3}\cdot \binom{15}{4}=\dfrac{15!}{3!12!}\cdot\dfrac{15!}{4!11!}=\dfrac{13\cdot14\cdot15}{2\cdot3}\cdot\dfrac{12\cdot13\cdot14\cdot15}{2\cdot3\cdot4}=13\cdot7\cdot5\cdot13\cdot7\cdot15=621075\\\\P(A)=\dfrac{621075}{2035800}=\dfrac{637}{2088}\approx30.5\%$

b)

$\displaystyle\\|A|=\binom{10}{7}\cdot 3^7=\dfrac{10!}{7!3!}\cdot2187=\dfrac{8\cdot9\cdot10}{2\cdot3}\cdot2187=4\cdot3\cdot10\cdot2187=262440\\\\P(A)=\dfrac{262440}{2035800}=\dfrac{243}{1885}\approx12.9\%$