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