Suppose your firm has three potential investments. The investments are either sucessful or not. Suppose that...

Suppose your firm has three potential investments. The investments are either sucessful or not. Suppose that each investment has probability 1/2 of being successful.

a. What is the probability that the third investment is successful?

b. What is the probability that the third investment is successful, given that the three investments are either all successful or all not successful?

c. What is the probability that the third investment is sucessful, given that two of the three investments is successful?

d. Suppose now that the probability that investment 1 is successful is 0.845, the probability that investment 2 is successful is 0.5505, and the probability that investment 3 is successful is 0.4. Consider these two events: A: two of the three investments are successful, and B: investment 3 is successful. Are these events independent? Why or why not?

Solutions

Expert Solution

a)P(third investment is successful) =1/2 (since for each investment probability of being successful is fixed)

P(all three successful or all three unsuccessful) =(1/2)³+(1/2)³ =0.25

P(third investment is successful |all three successful or all three unsuccessful)

=P(third investment is successful)/P(third investment is successful |all three successful or all three unsuccessful)

=(1/2)/0.25 =0.5

P(2 of three are success) =P(1st success and 2nd success and 3rd not success)+P(1st success and 2nd not success and 3rd success)+P(1st not success and 2nd success and 3rd success)

=(1/2)*(1/2)*(1/2)+(1/2)*(1/2)*(1/2)+(1/2)*(1/2)*(1/2)

=0.375

P(2 of three are success and 3rd success)= P(1st success and 2nd not success and 3rd success)+P(1st not success and 2nd success and 3rd success)

=(1/2)*(1/2)*(1/2)+(1/2)*(1/2)*(1/2) =0.25

therefore P( third investment is sucessful, given that two of the three investments is successful )

=0.25/0.375 =2/3

P(A) =P(2 of three are success) =P(1st success and 2nd success and 3rd not success)+P(1st success and 2nd not success and 3rd success)+P(1st not success and 2nd success and 3rd success)

=0.845*0.5505*(1-0.4)+0.845*(1-0.5505)*0.4+(1-0.845)*0.5505*0.4=0.465166

P(A n B)= P(1st success and 2nd not success and 3rd success)+P(1st not success and 2nd success and 3rd success)

=0.845*(1-0.5505)*0.4+(1-0.845)*0.5505*0.4 =0.186062

P(A)*P(B) =0.465166*0.4 =0.186066

since P(A)*P(B) is not equal to P(A n B) , therefore A and B are not independent

orchestra answered 1 year ago

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