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?

Updated On: Jul 5, 2022
  • $\frac{3}{204}$
  • $\frac{7}{408}$
  • $\frac{5}{204}$
  • $\frac{5}{408}$
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The Correct Option is D

Solution and Explanation

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