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Two insurance schemes A and B are to be sold An agent has the chance of finding customers as 60 percent for A and 40 percent for B Assuming the sale of A and B is independent find the probability.?
Wiki User
∙ 13y ago
P(X|Y) = P(Y intersection X) / P(Y); where
P(X|Y) is probability of event X provided event Y had already occurred
P(Y) is probability of event Y happening
P(Y intersection X) is probability of events Y & X occurring together
Q3.a
P(Y): Prob of at least one insurance schemes (A or B) has been sold
= 1 - Prob of none of the schemes sold
= 1 - (1-Prob of A being Sold)*(1-Prob of B being Sold) // As schemes A & B are independent
= 1 - (1-0.6)(1-0.4)
= 1 - (0.4)*(0.6)
= 1 - 0.24
= 0.76
P(X intersection Y): Prob of at least one insurance schemes (A or B) has been sold AND also scheme 'A' being sold
= Prob(A sold) and Prob (B sold) + Prob(A sold) and Prob (B not sold)
= 0.6 * 0.4 + 0.6 * (1-0.4)
= 0.24 + 0.36
= 0.6
P(X|Y): Prob scheme A has been sold given that at least one insurance scheme has been sold
= P(X intersection Y) / P(Y)
= 0.6 / 0.76
= 15/19
Wiki User
∙ 13y ago
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