Question

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

Answer 1

Let $R=\left\{5\text{red marbles}\right\}$ and $B=\left(3\text{black marbles}\right\}$

For at least one of the three marbles drawn be black, if the first marble is red, then thefollowing three conditions will be followed.
(i) Second marble is black and third is red $\left(E_1\right)$

(ii) Second marble is black and third is also black $\left(E_2\right)$
(iii) Second marble is red and third is black $\left(E_3\right)$

$P\left(E_1\right)=P\left(R_1\right)\cdot P\left(B_1/R_1\right)\cdot P\left(R_2/R_1{B}_1\right)=\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{4}{6}=\frac{60}{336}=\frac{5}{28}$
$P\left(E_2\right)=P\left(R_1\right)\cdot P\left(B_1/R_1\right)\cdot P\left(B_2/R_1{B}_1\right)=\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{2}{6}=\frac{30}{336}=\frac{5}{56}$
And $P\left(E_3\right)=P\left(R_1\right)\cdot P\left(R_2/R_1\right)\cdot P\left(B_1/R_1{R}_2\right)=\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}=\frac{60}{336}=\frac{5}{28}$
$\therefore P\left(E\right)=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=\frac{5}{28}+\frac{5}{56}+\frac{5}{28}$
$=\frac{10+5+10}{56}=\frac{25}{56}$