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In an urn there are 3 black, 4 white and 5 red balls. One chooses three balls without replacing.

a) Determine the probability that no ball is red.

b) Determine the probability that no ball is red given that one knows that all three have the same colours.

A) I solve the problem with (7/12)(6/11)(5/10)=0.159=15.9%

B) P(B|A)=(no red|all three chooses are the same colour).

P(B|A)=P(B intersect A)/ P(A)

P(A)=((3 over 3)(4 over 3)(5 over 3))/(12 over 3)= 0.06818

How do I calculate P(B intersect A)? I assume that they are dependent of each other and in that case I can´t write P(A intersect B)=P(A)*P(B), correct?

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    $\begingroup$ Consider a different solution strategy for (b): by deriving a formula for all three to have the same color conditional on the color, you can easily find the relative chances that all three are black, all three are white, and all three are red. The answer follows immediately from that. $\endgroup$
    – whuber
    Nov 12, 2019 at 16:25
  • $\begingroup$ Please add the self-study tag. $\endgroup$ Nov 20, 2019 at 3:06

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For the intersection we can simply write the following:

$$\begin{align}P(B\cap A)&=P(\text{no red }\cap \text{ all three same color})\\&=P(\text{all three green})+P(\text{all three white})\end{align}$$

I'm not sure about how you calculate your $P(A)$ by the way.

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