MHT CET · Maths · Probability
An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability, that the three balls have different colours, is
- A \(\frac{1}{3}\)
- B \(\frac{2}{7}\)
- C \(\frac{1}{21}\)
- D \(\frac{2}{23}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{7}\)
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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