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Bag I has 7 red and 3 green balls. Bag II has 3 red and 17 green.

(a) If the ball selected was a green, then what is the probability that Bag I was selected, and what is the probability that Bag II was selected?

(b) If a second ball I choose from the same bag with first time, what is the probability that the ball will be green; what is the probability that the ball will be red?

Remark: The ball first time pick will put back into bag.

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  • $\begingroup$ Is the first ball put back into the bag, before taking the second ball? $\endgroup$
    – Matti P.
    Commented Sep 10, 2019 at 5:18
  • $\begingroup$ I edit the question, thx $\endgroup$
    – Wendy
    Commented Sep 10, 2019 at 5:25
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    $\begingroup$ I think I smell Bayes' formula: $$P(A|B)P(B) = P(B|A)P(A) $$ $\endgroup$
    – AlvinL
    Commented Sep 10, 2019 at 5:41
  • $\begingroup$ Writing the probabilities in math terms, we get $$ P(\text{red}|~\text{I}~) = \frac{7}{10} \qquad P(\text{green}|~\text{I}~) = \frac{3}{10} $$ $$ P(\text{red}|~\text{II}~) = \frac{3}{20} \qquad P(\text{green}|~\text{II}~) = \frac{17}{20} $$ $\endgroup$
    – Matti P.
    Commented Sep 10, 2019 at 5:48
  • $\begingroup$ Is there a general probability for choosing Bag I ? Like "a priori"-probability, so that we are not given any other information? $\endgroup$
    – Matti P.
    Commented Sep 10, 2019 at 5:55

2 Answers 2

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a)First, since there 2 times more balls in bag II you can see that repeating this experiment many times, the balls from bag I are gonna be pick twice as much.The balls from bag one count double then.

You know you got a green ball, and there is 23 green balls in total (6=3*2 in bag I because they count double) + (17 in bag II):

  • there is 6 out of 23 chances to pick from bag I = 6/23
  • 17 out of 23 chances to pick from bag II = 17/23

b) In this case you know you gotta pick from only one bag, you chances to pick red or green are the number of green/red ball out of the total number of balls in that unique bag.

  • bag I : 7/10 chances to pick red and 3/10 chances to pick green
  • bag II : 3/20 chances to pick red and 17/20 chances to pick green

this is an intuitive answer, for the exact formula you can look the previous comments on your question.

Have a nice day.

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  • $\begingroup$ No. The green balls do not all have an equal probability weighting. There are twice as many balls in the second bag, so one each is half as probable to be selected as a ball in the first bag (assuming the bags were selected without bias). $\endgroup$
    – Graham Kemp
    Commented Sep 10, 2019 at 11:11
  • $\begingroup$ Thanks you for pointing that out, It's was clearly a bad mistake from my part.It is not that easy to see this with the intuition and I was to fast, that is why the formulas are always useful. $\endgroup$
    – rapasite
    Commented Sep 10, 2019 at 12:31
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Bag I has 7 red and 3 green balls. Bag II has 3 red and 17 green.

(a) If the ball selected was a green, then what is the probability that Bag I was selected, and what is the probability that Bag II was selected?

Let $G_1, G_2$ be the event of selecting a green ball in the first or second draw, and $B_{\rm I}, B_{\rm II}$ be the event of selecting the respective bag.

Then you may use Bayes' Rule (and the Law of Total Probability):$$\mathsf P(B_{\rm I}\mid G_1)=\dfrac{\mathsf P(B_{\rm I})~\mathsf P(G_1\mid B_{\rm I})}{\mathsf P(B_{\rm I})~\mathsf P(G_1\mid B_{\rm I})+\mathsf P(B_{\rm II})~\mathsf P(G_1\mid B_{\rm II})}$$

Likewise you can fine $\mathsf P(B_{\rm II}\mid G_1)$.

Alternatively, first principle reasonings about the probability weight of each individual green ball will suffice (each from the ten balls in the first bag are twice as likely to have been picked as each from the twenty balls in the second bag, and so when given that a green ball was selected, there the odds of it coming from the first bag (with three green balls) rather than the second (with seventeen green balls) is ...).

(b) If a second ball I choose from the same bag with first time, what is the probability that the ball will be green; what is the probability that the ball will be red?

Assuming you are seeking the conditional probability when given the first draw was green, then just use the Law of Total Probability.

$$\mathsf P(G_2\mid G_1)=\mathsf P(G_2\mid B_{\rm I},G_1)~\mathsf P(B_{\rm I}\mid G_1)+\mathsf P(G_2\mid B_{\rm II},G_1)~\mathsf P(B_{\rm II}\mid G_1)$$

Further, if the draws are with replacement, then the draws are conditionally independent for a given bag, and things simplify:

$$\mathsf P(G_2\mid G_1)=\mathsf P(G_2\mid B_{\rm I})~\mathsf P(B_{\rm I}\mid G_1)+\mathsf P(G_2\mid B_{\rm II})~\mathsf P(B_{\rm II}\mid G_1)$$

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