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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?
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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