A bag contains 2 white, 3 green and 5 red balls. If three balls are drawn one after the other without…
A bag contains 2 white, 3 green and 5 red balls. If three balls are drawn one after the other without replacement, then the probability that the last ball drawn was red is
$\frac{2}{3}$
$\frac{3}{4}$
$\frac{5}{9}$
$\frac{1}{2}$
Solution
Since, A bag has 2 white, 3 green and 5 red balls.
Now, number of ways of choosing 3 ball such that last ball is $\mathrm{red}=8 \times 9 \times 5$
So, required probability $=\frac{8 \times 9 \times 5}{10 \times 9 \times 8}=\frac{1}{2}$.