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}=(30!)/(7!23!)=(24\cdot25\cdot\ldots\cdot30)/(2\cdot3\cdot\ldots\cdot 7)=2035800`

a)

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

b)

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