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Suppose the events B1,B2,B3 are mutually exclusive and complementary events, such thatP(B1)=0.2, P(B2)=0.4and P(B3)=0.5. Consider another event A such thatP(AB1)=P(AB2)=0.1andP(AB3)=0.2Use Baye鈥檚 Rule to find

a.P(B1A)

b.PB2A

c.role="math" localid="1658214716845" P(B3A)

Short Answer

Expert verified

Therefore, the values of all parts are:

  1. 0.158
  2. 0.073
  3. 0.768

Step by step solution

01

Important formula

Baye鈥檚 formula is used for finding the conditional probability of the events.

The required formula is

PBiA=P(BiA)P(A)=P(Bi)P(ABi)P(B1)P(AB1)+P(B2)P(AB2)+...+P(Bk)P(ABk)

02

(a) Find the value of  P(B1A)

Here P(AB1)=P(AB2)=0.1, andP(AB3)=0.2.

Apply the baye鈥檚 formula, then

PB1A=P(B1)P(AB1)P(B1)PAB1+P(B2)P(AB2)+P(B3)P(AB3)=(0.2)(0.4)(0.2)(0.4)+(0.15)(0.25)+(0.65)(0.6)=0.158

So, the values of all parts are 0.158

03

(b) Evaluate the value of P(B2A)

PB2A=P(B2)P(AB2)P(B1)P(AB1)+P(B2)P(AB2)+P(B3)P(AB3)=(0.15)(0.25)(0.2)(0.4)+(0.15)(0.25)+(0.65)(0.6)=0.073

Hence, the values of all parts are 0.073

04

(c) Determine the value of P(B3A)

PB3A=P(B3)P(AB3)P(B1)PAB1+P(B2)PAB2+P(B3)P(AB3)=(0.65)(0.6)(0.2)(0.4)+(0.15)(0.25)+(0.65)(0.6)=0.768

Therefore, the values of all parts are 0.768

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