0
$\begingroup$

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?

MyApproach:

Case1: 1 white ball from bag 1 and 2 balls from bag 2 . So probability is $C^5_1*{C^9_2}/{C^9_1*{C^{16}_2}}$

Case2: I can't understand how to solve this part.

$\endgroup$
  • $\begingroup$ So the bags are chosen randomly as well? $\endgroup$ – Henno Brandsma Feb 13 '18 at 12:52
  • $\begingroup$ @HennoBrandsma I don't know.May be. $\endgroup$ – user517784 Feb 13 '18 at 12:53
  • $\begingroup$ @HennoBrandsma wait you are right.Yes i have mentioned that in case 1. $\endgroup$ – user517784 Feb 13 '18 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. $\endgroup$ – Henno Brandsma Feb 13 '18 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 Feb 13 '18 at 12:58
1
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy