SOLUTION: Carlee (again!) is spinning a spinner marked with the letters a through f . She defines the events A={a,b,c,d} and B={c,d,e} . Suppose that P(a)=0.02 , P(b)=P(c)=0.1 , P(d)=0.11

Algebra ->  Permutations -> SOLUTION: Carlee (again!) is spinning a spinner marked with the letters a through f . She defines the events A={a,b,c,d} and B={c,d,e} . Suppose that P(a)=0.02 , P(b)=P(c)=0.1 , P(d)=0.11       Log On


   



Question 1031842: Carlee (again!) is spinning a spinner marked with the letters a through f . She defines the events A={a,b,c,d} and B={c,d,e} . Suppose that P(a)=0.02 , P(b)=P(c)=0.1 , P(d)=0.11 , and P(e)=0.04 .
(a) P(A)=
(b) P(B)=
(c) P(A∩B)=
(d) P(A|B)=
(e) P(B|A)=
(f) Are A and B independent? Your answer should be "Y" or "N".

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Carlee (again!) is spinning a spinner marked with the letters a through f . She defines the events A={a,b,c,d} and B={c,d,e} .
Suppose that
P(a)=0.02 ,
P(b)=P(c)=0.1 ,
P(d)=0.11 ,
and P(e)=0.04 .
(a) P(A)= 0.02+0.10 + 0.10 + 0.11 = 0.33
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(b) P(B)= 0.10 + 0.11 + 0.04 = 0.25
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(c) P(A∩B)= P{c,d} = 0.10+0.11 = 0.21
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(d) P(A|B)= P(A and B)/P(B) = 0.21/0.25
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(e) P(B|A)= P(A and B)/P(A) = 0.21/0.33
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(f) Are A and B independent? Your answer should be "Y" or "N".
Question:: P(A|B) = P(A)*P(B)
0.21/0.25 = 0.33*0.25
0.84 = 0.0825
Ans:: not equal
Ans: A and B are not independent
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Cheers,
Stan H.
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