A bag contains 10 similar balls, of which 4 are blue and 6 are red. Three balls are taken out at random from…
A bag contains 10 similar balls, of which 4 are blue and 6 are red. Three balls are taken out at random from the bag one after the other without replacement. The probabilities that all the three balls drawn are red is
$\frac{1}{5}$
$\frac{1}{6}$
$\frac{5}{9}$
$\frac{1}{2}$
Solution
Total no. of balls $=10$
No, of blue balls $=4$
No. of red balls $=6$
No. of ways to draw three red balls $={ }^6 \mathrm{C}_3$
$\therefore$ The required probability is :
$
\mathrm{P}=\frac{{ }^6 \mathrm{C}_3}{10_{\mathrm{C}_3}}=\frac{\frac{6 \times 5 \times 4}{6}}{\frac{10 \times 9 \times 8}{6}}=\frac{1}{6}
$