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Box A has 3 red and 7 white, and Box B has 8 red and 2 white.

What is the probability that Box B is selected given 2 red balls are drawn? i.e. $$\Pr(\text{ Box B}\mid\text{2 red balls }). $$

I understand selecting Box B randomly = 50% chance (1/2), selecting 1st red ball = 80% chance(8/10), 2nd red ball = 77% chance(7/9), but I don't know how to put it all together and I feel like I'm missing something.

appreciate the help.

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    $\begingroup$ Pr(Box B|2 red balls) means "the probability that Box B is selected given that 2 red balls are drawn" not "the probability that Box B is selected and 2 red balls are drawn", Which one do you mean? $\endgroup$
    – Alan Abraham
    Commented Sep 2, 2021 at 1:32
  • $\begingroup$ excuse me, I meant to say the 1st phrase "The probability that Box B is selected given that 2 red balls are drawn" computing Pr(Box B| 2 red balls). Can someone show me the product rule for this? When I try it I seem to be off. I've been told it's meant to equal 28/31 $\endgroup$
    – Alan Greene
    Commented Sep 2, 2021 at 2:58
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    $\begingroup$ Firstly correct your question in the body to show what exactly you are asking $\endgroup$
    – true blue anil
    Commented Sep 2, 2021 at 4:33

1 Answer 1

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Your probability is

$$\mathbb{P}(B|RR)=\frac{\frac{8}{10}\cdot\frac{7}{9}}{\frac{3}{10}\cdot\frac{2}{9}+\frac{8}{10}\cdot\frac{7}{9}}=\frac{56}{62}=\frac{28}{31}$$

as you expected to find

the chance of 50% selecting A or B is irrelevant being the same for the two boxes

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  • $\begingroup$ Thank you!! I swear I tried this, but I must have miscalculated something I was only getting .83 I was loosing my mind $\endgroup$
    – Alan Greene
    Commented Sep 2, 2021 at 15:21

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