anonymous
  • anonymous
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 P(FIE)=??
Mathematics
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anonymous
  • anonymous
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 P(FIE)=??
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
P(F|E)=P(F intersection E)/P(E) P(E+F)= (18/68)*(50/67) P(E)=18/68 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
anonymous
  • anonymous
would that be reduced to 25/64
anonymous
  • anonymous
No, Andras is saying the answer would be 50/67, with which I agree. :)

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anonymous
  • anonymous
how did you get 25/64?
anonymous
  • anonymous
you wrote 50/68
anonymous
  • anonymous
that is P(F) on its own, P(F|E)=50/67
anonymous
  • anonymous
do you know what is the difference?
anonymous
  • anonymous
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
anonymous
  • anonymous
You can do the work here :) I will double check it for you, is that fine?
anonymous
  • anonymous
you need to count the probability that they occur together and the probability of P(E)

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