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 $P (C |A \cap B ) =\frac{3}{16}$
$\therefore P(A \cap B \cap C) = P(A).P(B|A).P(C|A \cap B)$
$=\frac{5}{18} \times \frac{4}{17} \times \frac{3}{16}$
$= \frac{5}{408}$