Question

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)=??

Answer 1

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

Answer 2

would that be reduced to 25/64

Answer 3

No, Andras is saying the answer would be 50/67, with which I agree. :)

Answer 4

how did you get 25/64?

Answer 5

you wrote 50/68

Answer 6

that is P(F) on its own, P(F|E)=50/67

Answer 7

do you know what is the difference?

Answer 8

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

Answer 9

You can do the work here :) I will double check it for you, is that fine?

Answer 10

you need to count the probability that they occur together and the probability of P(E)