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A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. From the box, what is the probability that:(i) all are blue? (ii) at least one is green? |
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Answer» Given: box containing 10 red, 20 blue and 30 green marbles. Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) five marbles are drawn from the given box, total possible outcomes are 60C5 therefore n(S) = 60C5 (i) let E be the event of getting all blue balls n(E)= 20C5 P(E) = \(\frac{n(E)}{n(S)}\) P(E) = \(\frac{^{20}C_5}{^{60}C_5}\) = \(\frac{34}{11977}\) (ii) let E be the event of getting at least one green let E’ be the event of getting no green E’= {(5B) (1R 4B) (2R 3B) (3R 2B) (4R 1B) (5R)} 5B = 20C5 1R 4B = 10C1 20C4 2R 3B = 10C2 20C3 3R 2B = 10C3 20C2 4R 1B= 10C4 20C1 5R= 10C5 P(E) =1 - P(E’) P(E’) = \(\frac{^{20}C_5+^{10}C_1^{20}C_4+^{10}C_2^{20}C_3+^{10}C_3^{20}C_2+^{10}C_4^{20}C_1+^{10}C_5}{^{60}C_5}\) P(E’) = \(\frac{117}{4484}\) P(E) = 1 - \(\frac{117}{4484}\) = \(\frac{4367}{4484}\) |
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