⁦Probability and combinations⁩

⁦Find the probability of an event with P(A) = m : n when the outcomes of a choice are counted with combinations Cₙ⁩ᵏ ⁦and arrangements Aₙ⁩ᵏ.

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الدليل

⁦Topic from the Ukrainian Grade 11 textbook (algebra and calculus with geometry), section 3⁩ "⁦Elements of combinatorics, probability and statistics⁩", ⁦item 14⁩ "⁦The classical definition of the probability of a random event⁩". ⁦When all outcomes of an experiment are equally likely, P(A) = m : n, where n is the number of all possible outcomes and m the number of outcomes favourable to A⁩.

⁦When k of n objects are chosen at random and the order does not matter, n = Cₙ⁩ᵏ = n! : (⁦(n − k)⁩! · k!). ⁦When objects are chosen from two groups at once, the number of favourable outcomes m is the product of the numbers of ways to choose from each group⁩. ⁦When the order matters (a number with different digits), use arrangements Aₙ⁩ᵏ = n! : ⁦(n − k)⁩!. ⁦Solutions are laid out as in the book⁩: "Given", "Find", "Solution", "Answer". ⁦The answer is a fraction in lowest terms⁩.

⁦The textbook example (problem 3)⁩: ⁦of 28 apples, 15 are yellow and 3 apples are chosen; P = C₁₅³ : C₂₈³ = 5 : 36⁩.