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Home » An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls drawn in the first draw are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and the second draw gives 3 black balls.
An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls drawn in the first draw are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and the second draw gives 3 black balls.
CBSE | Class 12 | Exercise 29B | Maths | Probability | RS Aggarwal
An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls drawn in the first draw are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and the second draw gives 3 black balls.

Let, success in the first draw be getting 3 white balls.
Now, the Probability of success in the first trial is
P_{1}(\text { success })=\frac{5_{c_{3}}}{13_{c_{3}}}=\frac{10}{286}=\frac{5}{143}
Let success in the second draw be getting 3 black balls.
Probability of success in the second trial without replacement of the first draw is given by
P_{2} \text { (success) }=\frac{8_{c_{3}}}{10_{c_{3}}}=\frac{56}{120}=\frac{7}{15}
Hence, the probability that the first draws gives 3 white and the second draw gives 3 black balls, with each trial being independent is given by
\mathrm{P}_{1} \times \mathrm{P}_{2}=\frac{5}{143} \times \frac{7}{15}=\frac{7}{429}

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