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An urn contains 4 red and 5 white balls. Two balls are drawn one after the other
without replacement, then the probability that both the balls are red is
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without replacement, then the probability that both the balls are red is
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The correct answer is:
$\frac{1}{6}$
Red balls $=4$ and White balls $=5 \Rightarrow$ Total balls $=4+5=9$ Two balls are drawn one after the other without replacement
$n(S)={ }^{9} C_{1} \times{ }^{8} C_{1}=9 \times 8$
$\text { Required probability }=\frac{{ }^{4} C_{1} \times{ }^{3} C_{1}}{9 \times 8}=\frac{4 \times 3}{9 \times 8}=\frac{1}{3 \times 2}=\frac{1}{6}$
$n(S)={ }^{9} C_{1} \times{ }^{8} C_{1}=9 \times 8$
$\text { Required probability }=\frac{{ }^{4} C_{1} \times{ }^{3} C_{1}}{9 \times 8}=\frac{4 \times 3}{9 \times 8}=\frac{1}{3 \times 2}=\frac{1}{6}$
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