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Question:

A bag contains $5$ white and $4$ black balls and another bag contains $7$ white and $9$ black balls . A ball is drawn from the first bag and two balls are drawn from the second bag. What is the probability of drawing one white and two black balls?

My Approach:

Case $1$: 1 white ball from bag $1$ and $2$ balls from bag $2$ . So probability is $$\frac{\binom{5}{1}\times \binom{9}{2}}{\binom{9}{1}\times\binom{16}{2}}$$

Case $2$: I can't understand how to solve this part.

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  • $\begingroup$ So the bags are chosen randomly as well? $\endgroup$ Commented Feb 13, 2018 at 12:52
  • $\begingroup$ @HennoBrandsma I don't know.May be. $\endgroup$
    – user517784
    Commented Feb 13, 2018 at 12:53
  • $\begingroup$ @HennoBrandsma wait you are right.Yes i have mentioned that in case 1. $\endgroup$
    – user517784
    Commented Feb 13, 2018 at 12:54
  • $\begingroup$ So the other case is 1 ball from the second bag (7/9) and 2 from the other? Then take the average of these 2 probabilities. $\endgroup$ Commented Feb 13, 2018 at 12:56
  • $\begingroup$ @HennoBrandsma case 2 should be I think 1 black ball from bag 1 and 1 white ball and 1 black ball from bag 2. $\endgroup$
    – user517784
    Commented Feb 13, 2018 at 12:58

1 Answer 1

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Let's draw from the first bag at first.

Then:

$$P(wbb)+P(bbw)+P(bwb)=\frac59\frac9{16}\frac8{15}+\frac49\frac9{16}\frac7{15}+\frac49\frac7{16}\frac9{15}$$

where e.g. $bwb$ stands for the event that at first a black ball is drawn (from the first bag) secondly a white ball (from the second bag) and thirdly a black ball (from the second bag).

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  • $\begingroup$ The answer given in my book is 1/6(case1)+7/30(case2)=2/5. $\endgroup$
    – user517784
    Commented Feb 13, 2018 at 13:08
  • $\begingroup$ Can you just check it once. $\endgroup$
    – user517784
    Commented Feb 13, 2018 at 13:08
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    $\begingroup$ That's also my answer (as you can check yourself). $\endgroup$
    – drhab
    Commented Feb 13, 2018 at 13:14
  • $\begingroup$ Yes thanks for answering $\endgroup$
    – user517784
    Commented Feb 13, 2018 at 13:15
  • $\begingroup$ You are welcome. $\endgroup$
    – drhab
    Commented Feb 13, 2018 at 13:18

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