Question 1: Marks:4+2+2=8 a) The probability that atrainee will remain with a co

Question

Question 1:                                                                                                                     Marks:4+2+2=8

a)    The probability that atrainee will remain with a company is 0.6. The probability that anemployee earns more than Rs.10, 000 per year is 0.5. Theprobability that an employee is a trainee who remained with thecompany or who earns more than Rs.10, 000 per year is 0.7. What isthe probability that an employee earns more than Rs.10, 000 peryear given that he is a trainee who stayed with the company.

b)    The odds against student Xsolving a Business statistics problem are 8 to 6, and odds infavour of student Y solving the problem are 14 to 16.

Explanation / Answer

a) Let A denote the event thatthe trainee will remain with the company. Given P(A) = 0.6 Let B denote the event that an employee earns more thanRs.10,000 per year. Given P(B) = 0.5 The probability that an employee is a trainee whoremained with the company or who earns more thanRs.10,000 per year is given by the probability : P(AUB)     So, P(AUB) = 0.7 (given)    P(AB) = P(A) +P(B) -P(AUB)                 =0.6 +0.5 -0.7                 = 0.4 The probability that an employee earns more thanRs.10,000 per year given that he is a trainee whostayed with the company is given by the conditional probability:     P(B/A) =P(AB)/P(A)                = 0.4/0.6                =0.66 b) Let A denote the event that studentX solves a problem.     The odds against X solving the problem are8 to 6. P(A) = 6/(8+6)             =6/14             =3/7     Let B denote the event that student Ysolves a problem.     The odds in favour of student Y solving theproblem are 14 to 16. P(B) = 14/(14+16)           =14/30           =7/15 The chance that the problem will be solved if they bothtry independent of each other = P(AB)     P(AB) = P(A).P(B)   (since they try independently the two events are independent)                  =(3/7).(7/15)                  =0.2     The probability that none of them is ableto solve the problem is given by :        1 - Prbability thatatleast one of them will solve the problem =1 - P(AUB)                                                                                                  =1-[P(A)+P(B)-P(AB)]                                                                                                  =1-[3/7 + 7/15 - (3/7).(7/15)]                                                                                                   =1- 0.6952                                                                                                   =0.3048

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