Conditional Probability : S1 Edexcel January 2011 Q7f : ExamSolutions Maths Revision

[darren]

I don't get the part where A intersection B changes to just red

[camillecandy]

how did the probability A intersection B become just probability A as in how did it become the probability of all red rather than probability of all the colours red intersection all the same colours?

[sadmanzaid]

I feel if people knew bayes' theorem of probability this question would've been a piece of cake, so let me impart to those needed: P(A|B) = P(A∪B)/P(B) also P(B|A) = P(A∪B)/P(A) and hence P(A|B) = P(B|A)•P(A)/P(B).

Here P(A|B) is probability of getting all red given that all three balls drawn have the same color so what's the reverse? Probability of getting all three balls of the same color given that they are all red which is obviously 1 therefore the formula results in P(A|B) = P(B|A)•P(A)/P(B) = 1•P(A)/P(B) which is just the probability of all balls being red divided by probability of all three having the same color.

[mindfulawareness]

Hard one. Newbie, but why not just 1/2*2/5*2/3? That's what the tree diagram shows. Not disputing obviously, but as a learner there's some non-visual counter-intuitiveness to be overcome 😊. ...and that is overcome when this is illustrated with a Venn Diagram 😊
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