MHT CET · Maths · Three Dimensional Geometry
An urn contains 9 balls of which 3 are red, 4 are blue and 2 are green. Three balls are drawn at random from the urn. The probability that the three balls have difference colours is
- A \(\frac{1}{14}\)
- B \(\frac{3}{14}\)
- C \(\frac{1}{7}\)
- D \(\frac{2}{7}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{7}\)
Step-by-step Solution
Detailed explanation
\(\text { Required probability }=\frac{{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^9 \mathrm{C}_3}\)
\(=\frac{3 \times 4 \times 2}{\left(\frac{9 !}{3 ! 6 !}\right)}=\frac{24 \times 6}{9 \times 8 \times 7}=\frac{2}{7}\)
\(=\frac{3 \times 4 \times 2}{\left(\frac{9 !}{3 ! 6 !}\right)}=\frac{24 \times 6}{9 \times 8 \times 7}=\frac{2}{7}\)
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