a container contains 50 green tokens, 15 blue tokens, and 3 red tokens. two tokens are randomly selected without replacement compute P(FIE).
E-you select a non green token first
F-the second token is green
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P(F|E)=P(F intersection E)/P(E)
So P(F|E)=(18/68)*(50/67) /(18/68)= 50/67
P(F) would be 50/68, so P(F|E) is a bit more likely
would that be reduced to 25/64
No, Andras is saying the answer would be 50/67, with which I agree. :)
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how did you get 25/64?
you wrote 50/68
that is P(F) on its own, P(F|E)=50/67
do you know what is the difference?
so what would the P(FIE) be for this problem then
a container contains 30 green tokens ,20 blue tokens, and 4 red tokens. two tokens are randomly selected without replacement compute P(FIE).
E-you select a non red token first
F- the select token is red
You can do the work here :) I will double check it for you, is that fine?
you need to count the probability that they occur together and the probability of P(E)