# Probability of drawing $5$ white and $4$ black balls and another bag of $7$ white and $9$ black balls.

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.

• So the bags are chosen randomly as well? Feb 13, 2018 at 12:52
• @HennoBrandsma I don't know.May be.
– user517784
Feb 13, 2018 at 12:53
• @HennoBrandsma wait you are right.Yes i have mentioned that in case 1.
– user517784
Feb 13, 2018 at 12:54
• 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. Feb 13, 2018 at 12:56
• @HennoBrandsma case 2 should be I think 1 black ball from bag 1 and 1 white ball and 1 black ball from bag 2.
– user517784
Feb 13, 2018 at 12:58

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).

• The answer given in my book is 1/6(case1)+7/30(case2)=2/5.
– user517784
Feb 13, 2018 at 13:08
• Can you just check it once.
– user517784
Feb 13, 2018 at 13:08
• That's also my answer (as you can check yourself). Feb 13, 2018 at 13:14