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.
Solution
Correct Answer: Option B
Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then
Required probability = \(P\left(A\cap B\cap C\right)\)
= \(P\left(A\right)\;P\left(\frac BA\right)\;P\left(\frac C{A\cap B}\right)\)
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
= \(\therefore P\left(\frac BA\right)=\frac3{17}\)
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.
\(\therefore P\left(\frac C{A\cap B}\right)\;=\frac2{16}=\frac18\)
Hence the required probability = \(\frac29\times\frac3{17}\times\frac18=\frac1{204}\)
Practice More Questions on Our App!
Download our app for free and access thousands of MCQ questions with detailed solutions