Question

Probablity question

Answer 1

Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) (B) P(A′B′) = [1 – P(A)] [1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1

Answer 2

i don't see the correct answer there

Answer 3

independent means \[P(A|B)=P(A)\]

Answer 4

?

Answer 5

the answer is there

Answer 6

or
\[P(A\cap B)=P(A)P(B)\]
guess we need to look more carefully then huh?

Answer 7

ooh is that B just
\[P(A'\cap B')=(1-P(A))(1-P(B))\]?

Answer 8

yes

Answer 9

sorry got confused with the proliferation of B's there

Answer 10

@Zarkon can explain why this one is the correct one

Answer 11

well I assume that is what they meant

Answer 12

process of elimination makes that what they meant

Answer 13

I think demorgan's theorem would be helpful

Answer 14

\[A' \text{ and } B'=(A \text{ or } B)'\]

Answer 15

\[P((A \text{ or } B)')=1-P(A \text{ or } B) \]
then you can use that one thing for P(A or B)=P(A)+P(B)-P(A and B)

Answer 16

and misty said earlier P(A and B)=PA) P(B)
since A and B are independent

Answer 17

P(A′B′) = [1 – P(A)] [1 – P(B)]=1-P(A)-P(B)+P(A)P(B)


-1+P(A)+P(B)+P(A'B')=P(A)P(B)
-1+(1-P(A'))+(1-P(B'))+P(A'B')=P(A)P(B)
1-(P(A')+P(B')-P(A'B')=P(A)P(B)
1-P(A'\(\cup\)B')=P(A)P(B)
P(AB)=P(A)P(B)

Answer 18

i think you only need to know that
\[1-P(A)=P(A')\]

Answer 19

like old home week today!