An urn contains 4 white 6 black and 8 red… | Answer

Subject

Quantitative Aptitude

Topic

Probability

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.
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ক)

5/204

খ)

1/204

গ)

13/204

ঘ)

None of these

Explanation

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}

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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.Reading

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.

Explanation

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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.
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