Question

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

• A. $\frac{15}{32}$
• B. $\frac{23}{36}$
• C. $\frac{21}{37}$
• D. $\frac{25}{56}$

Answer 1

Answer: (D)

Let $A$ be the event that the first marble is red.
Let $B$ be the event that the second marble is black.
Let $C$ be the event that the second marble is red.
Let $D$ be the event that the third marble is black.
Let $E$ be the event that the third marble is red.
The required probability
$P(A)\times P(B/A)\times P(D/A\cap B)+P(A)\times P(C/A)$
$\times P(D/A\cap C)+P(A)\times P(B/A)\times P(E/A\cap B)$
$=\frac{5}{8}\times \frac{3}{7}\times \frac{2}{6}+\frac{5}{8}\times \frac{4}{7}\times \frac{3}{6}+\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}$
$=\frac{30}{336}+\frac{60}{336}+\frac{60}{336}=\frac{30+60+60}{336}=\frac{150}{336}=\frac{25}{56}$