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anonymous
 5 years ago
a container contains 50 green tokens, 15 blue tokens, and 3 red tokens. two tokens are randomly selected without replacement compute P(FIE).
Eyou select a non green token first
Fthe second token is green
P(FIE)=??
anonymous
 5 years ago
a container contains 50 green tokens, 15 blue tokens, and 3 red tokens. two tokens are randomly selected without replacement compute P(FIE). Eyou select a non green token first Fthe second token is green P(FIE)=??

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P(FE)=P(F intersection E)/P(E) P(E+F)= (18/68)*(50/67) P(E)=18/68 So P(FE)=(18/68)*(50/67) /(18/68)= 50/67 P(F) would be 50/68, so P(FE) is a bit more likely

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would that be reduced to 25/64

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, Andras is saying the answer would be 50/67, with which I agree. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get 25/64?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is P(F) on its own, P(FE)=50/67

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know what is the difference?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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). Eyou select a non red token first F the select token is red

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can do the work here :) I will double check it for you, is that fine?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to count the probability that they occur together and the probability of P(E)
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