# Porbability of selecting balls from boxes [closed]

There are three boxes. B1, B2, B3 The probability of selecting them is 0.2, 0.2 , 0.6 respectively.

B1 contains 3 red balls and 7 green balls. B2 contains 5 red balls and 5 green balls. B3 contains 2 red balls and 8 green balls.

If we select a box and then a ball from the box what is the probability that the ball is of red color.

If we select the a ball and it turns out to be of green color what is the probability that it comes from B3 ?

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## closed as too localized by Qiaochu Yuan, Robin Chapman, Yemon Choi, Victor Protsak, Harry GindiAug 21 '10 at 7:06

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MO is not really for such questions; have you read the FAQ? – Qiaochu Yuan Aug 21 '10 at 6:36

## 1 Answer

Well, this looks more like someone trying to get their homework done, but for the first part:

$p = 0.2 * \frac{3}{3+7} + 0.2 * \frac{5}{5+5} + 0.6 * \frac{2}{2+8}$

$p = 0.06 + 0.10 + 0.12$

$p = 0.28$

Showed the work for you too.

So if the probability that a chosen ball is red is 28%, then the probability that a chosen ball is green is 72%.

So what is {probability that chosen ball came from B3 | chosen ball is green}? Look up conditional probability, look up bayesian, etc.

$p_g = 0.2 * \frac{7}{3+7} + 0.2 * \frac{5}{5+5} + 0.6 * \frac{8}{2+8}$

$p_g = 0.14 + 0.10 + 0.48$

$p_g = 0.72$

{ $p_g3$ | green ball} = (picked from box 3 and green) / (picked green)

= 0.48 / 0.72

= 2 /3

Please do your homework yourself. Showed the work for you too.

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Oh no the problem I am solving is entirely different. Just wanted to check that my line of thinking is correct. – Akshar Prabhu Desai Aug 21 '10 at 6:50
-1 : as a matter of policy, if you think that something is homework then you shouldn't answer it. – Andy Putman Aug 21 '10 at 14:54
Andy, I'm sorry that I didn't know about the restriction on what to answer. Should I edit the answer away? – sleepless in beantown Aug 22 '10 at 0:01
@sleepless : At this point, editing it is kind of pointless. Just keep it in mind in the future (I certainly understand the urge to answer questions even if they seem too easy!). – Andy Putman Aug 24 '10 at 2:08