A bag contains 8 red and 4 green balls. Find the probability that two balls are red and one ball is green when three balls are drawn at random.
Solution
Correct Answer: Option B
\(n\left(S\right)=\;{}^{12}C_{4\;}=495\)
\(n\left(E\right)=\;{}^8C_2\times{}^4C_1=112\)
\(\therefore P(E)=\frac{112}{495}\)
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