A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (Binomial coefficients can be left as such)

Question

Aditi Chauhan · Answer

Hello Student,

Please find the answer to your question

Let us define the following events

A ≡ 4 white balls are drawn in first six draws

B ≡ 5 white balls are drawn in first six draws

C ≡ 6 white balls are drawn in first six draws

E ≡ exactly one white ball is drawn in next two draws (i.e. one white and one red)

Then P (E) = P (E| A) P (A) + P (E| B) P (B) + P (E| C) P (C)

But P (E |C) = 0 [As there are only 6 white balls in the bag.]

P (E) = P (E| A) P (A) + P (E| B) P (B)

Thanks
Aditi Chauhan
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A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (Binomial coefficients can be left as such)

A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (Binomial coefficients can be left as such)

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A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (Binomial coefficients can be left as such)

A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (Binomial coefficients can be left as such)

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