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so it P(EF)= .23/.6 AND P(FE)= .23/.40?

I have the formula EF=p(E AND F)/pF but I don't get how to work it.

Compute P(EF) AND P(FE) GIVEN P(E)=.4 P(F)=.6 AND P(E AND F)= .23

Compute P(EF) AND P(FE) GIVEN P(E)=.4 P(F)=.6 AND P(E AND F)= .23

I left out the f={1,11,21}. So would the answer be 21,26 or does it mean included in all E,F,G which would be none,because there is none that are in each one?

Let s={1,6,11,16,21,26}, E={6,16,26}, and G={21,26} Find the event E F AND G. Upside down u in between each letter. If there are no elements in the set, enter NONE

a wallet contains 8 $1 bills, 3 $5 bills and 8 $10 bills. A bill is selected at random from the wallet. Find themprobability of the following
a. the bill is a $1, given that it is smaller...
