Help with Probability Question9
Algebra

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Find the requested probability.
when 2 events A and B are indep , the occurrence of one event A doesnt affect the occurrence of the other event B and P(A intersection B) = P(A)*P(B)
Again, when A is indep of the occurrence of the other event B it goes without saying :A is indep of the nonoccurrence of B.i.e. A is indep of Bcap Hence P(A intersection Bcap = P(A)*P(Bcap)
These are the 2 rules we are going to make use of here.
P(AB) = P(A intersection B) /P(A) = P(A)*P(B) /P(A) = P(B) = 0.4
P(A \cup B) = P(A intersecton B cap)/ P(Bcap) = P(A)*P(B cap)/ P(Bcap) = P(A) = 0.6
third subdiv is the same as the first one
If you've meant P(B\cup A ) we ' ll use the same method specified above.
= P(B intersecton A cap)/ P(Acap) = P(B)*P(A cap)/ P(Acap) = P(B) = 0.4
b) A & B are dependent.so we CAN NOT use the rule P(A intersection B) = P(A)*P(B)
P(A/B) = P(AB)/P(B) = 0.3 (given)
so P(AB) / 0.4 = 0.3
P(AB) = 0.4*0.3 = 0.12
P(B/A ) = P(AB)/P(A) = 0.12 /0.6 = 0.2
P(A \cap B) = P(A intersection B cap) / P(Bcap) ..............Denote this by (M)
To find P(A intersection B cap) , consider (A intersection B cap ) in the Venn Diagram.
[That is you draw two circles interscting one another inside a rectangle.the 2 cicles correspond to the sets A & B . The rectangle represents the sample space]
You may notice (A intersection B cap) is A  AB
So P(A intersection B cap) = P( A  AB) = P(A)  P(AB) = 0.6  0.12 = 0.48
in the expression denoted by (M) we have obtained the numerator.
To get the denominator take P(Bcap) = P( non occurrence of B) = 1 P(B) = 1  0.4 = 0.6
So (M) becomes 0.48/0.6 =4/5 = 0.8
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