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Question
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 at least one is green?
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Solution
Out of 60 marbles, five marbles can be drawn in 60C5 ways.
∴ Total number of elementary events = 60C5
P (no green) = \[\frac{\text{ Favourable outcomes} }{\text{ Total outcomes } }\]
= \[\frac{{}^{30} C_5}{{}^{60} C_5}\]
Thus, P(at least one green) = 1 – P (no green)
\[= 1 - \frac{^{30}{}{C}_5}{^{60}{}{C}_5}\]
\[ = 1 - \frac{117}{4484}\]
\[ = \frac{4484 - 117}{4484}\]
\[ = \frac{4367}{4484}\]
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