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Quantitative Aptitude - Arithmetic Ability/
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Question

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

A) 5/204	B) 1/204
C) 13/204	D) None of these

Answer: B) 1/204

Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then

Required probability = $P (A \cap B \cap C)$

= $P (A) P (\frac{B}{A}) P (\frac{C}{A \cap B})$

Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

$∴ P (\frac{B}{A}) = \frac{3}{17}$

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

$∴ P (\frac{C}{A \cap B}) = \frac{2}{16} = \frac{1}{8}$

Hence the required probability = $\frac{2}{9} \times \frac{3}{17} \times \frac{1}{8} = \frac{1}{204}$

Subject: Probability

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