Question
A bag contains 5 red books, 7 yellow books
and 7 green books . 3 books are drawn randomly. What is the probability that the books drawn is not a green book ?Solution
Total number of books = 5 + 7 + 7 = 19 Let S be the sample space. Then, n(S) = number of ways of drawing 3 books out of 19 = 19C3 = `(19 xx 18 xx 17)/ (3 xx 2 xx 1)` = 969 Let E= event of drawing no green books n(E) = 12C3 = `(12 xx 11 xx 10)/(3 xx 2 xx 1)` = 220 `:.` P(E) = `(n(E))/(n(S))` = `220/969`
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