Question
A bag contains 10 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 = 10 + 7 + 7 = 24 Let S be the sample space. Then, n(S) = number of ways of drawing 3 books out of 24 = 24C3 = `(24 xx 23 xx 22)/ (3 xx 2 xx 1)` = 3036 Let E= event of drawing no green books n(E) = 17C3 = `(17 xx 16 xx 15)/(3 xx 2 xx 1)` = 680 `:.` P(E) = `(n(E))/(n(S))` = `680/3036` = `170/759`
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