Question

A bag contains $6$ red, $5$ blue and $7$ white balls. If three balls are drawn one by one (without replacement), then what is the probability that all three balls are blue?

• A. $\frac{3}{204}$
• B. $\frac{7}{408}$
• C. $\frac{5}{204}$
• D. $\frac{5}{408}$

Answer 1

Answer: (D)

Let $A$, $B$ and $C$ be events of drawing a red ball in first, second and third draw respectively. Then probability of getting red ball in all three draws
$P(A\cap B\cap C)=P(A).P(B|A).P(C|A\cap B)$
Now, $P(A)=\frac{5}{18}$, $P(B|A)=\frac{4}{17}$
and $P(C|A\cap B)=\frac{3}{16}$
$\therefore P(A\cap B\cap C)=P(A).P(B|A).P(C|A\cap B)$
$=\frac{5}{18}\times \frac{4}{17}\times \frac{3}{16}$
$=\frac{5}{408}$