EXAM P SAMPLE SOLUTIONS - soa.org

1 Page 1 of 91 SOCIETY OF ACTUARIES EXAM P PROBABILITY EXAM P SAMPLE SOLUTIONS Copyright 2018 by the Society of Actuaries Some of the questions in this study note are taken from past examinations. Some of the questions have been reformatted from previous versions of this note. Questions 154-55 were added in October 2014. Questions 156-206 were added January 2015. Questions 207-237 were added April 2015. Questions 238-240 were added May 2015. Questions 241-242 were added November 2015. Questions 243-326 were added September 2016.

2 Question 327 was added January 2018. Question 328 was added May 2018. Question 245 was deleted March 2020 Page 2 of 91 1. Solution: D Let G = viewer watched gymnastics, B = viewer watched baseball, S = viewer watched soccer. Then we want to find ()()( ) ( ) ( ) () () () ()()Pr1 = = + + + = ++ + = = 2. Solution: A Let R = referral to a specialist and L = lab work. Then [][][][][][]1[()][ ][ ] 1[] 1 L PR PL PR L PR PLP R LpR PLPRL = + = + + = + + = + + = 3.

3 Solution: D First note [] [ ] [ ] [][ ]cccPA B PA PB PA BPA BPA PBPA B = + = + Then add these two equations to get [][ ] [ ]()[]()[ ]()()[ ][ ][ ] ccPAB PABPA PB PBPAB PABPAPAB ABPAPAPA + = + + + + =+ =+ = Page 3 of 91 4. Solution: A For i = 1,2, let Ri = event that a red ball is drawn from urn i and let Bi = event that a blue ball is drawn from urn i. Then, if x is the number of blue balls in urn 2, () ()[][ ] [ ] [ ] [ ] Pr[] Pr[] PrPrPrPrPr416610161016RR BBRRBBRR BBxxx= = + =+ =+ ++ Therefore, xxxxx+=+=++ ++=+== 5.

4 Solution: D Let N(C) denote the number of policyholders in classification C. Then N(Young and Female and Single) = N(Young and Female) N(Young and Female and Married) = N(Young) N(Young and Male) [N(Young and Married) N(Young and Married and Male)] = 3000 1320 (1400 600) = 880. 6. Solution: B Let H = event that a death is due to heart disease F = event that at least one parent suffered from heart disease Then based on the medical records, 210 102108937937937 312625937937ccPH FPF == == and 108108625| 937625cccPH FPH FPF === = Page 4 of 91 7.

5 Solution: D Let A = event that a policyholder has an auto policy and H = event that a policyholder has a homeowners policy. Then, ()()( ) ()()( ) () HHAH = = = = = = = and the portion of policyholders that will renew at least one policy is given by ()()()( )( ) ( )( ) ( )( )() 53%ccAHA HAH + + =++== 8. Solution: D Let C = event that patient visits a chiropractor and T = event that patient visits a physical therapist. We are given that [ ] [ ]()() + = = Therefore, [] [ ] [ ] [][ ][ ][ ] 1 PrPrPrPrPr 2 TCTCTCTTTT = = = + = ++ = or [ ]() 2 += 9.

6 Solution: B Let M = event that customer insures more than one car and S = event that customer insurers a sports car. Then applying DeMorgan s Law, compute the desired probability as: ()()()( ) ( ) ()( ) ( )( )( )( )( )PrPr1 Pr1 PrPrPr1 PrPrPrPr1 SMSMSMSMSMSSMM = = = + = += += 10. This question duplicates Question 9 and has been deleted Page 5 of 91 11. Solution: B Let C = Event that a policyholder buys collision coverage and D = Event that a policyholder buys disability coverage.

7 Then we are given that [ ] 2 [ ] and [] D= =. By the independence of C and D, [][ ] [ ] 2 [ ][ ] / 2 [ ] , [ ] 2 D PCPDPDPDPDPC= ====== Independence of C and D implies independence of Cc and Dc. Then [][] [] (1 2 )( ) cPCDPC PD == = 12. Solution: E Boxed numbers in the table below were computed. High BP Low BP Norm BP Total Regular heartbeat Irregular heartbeat Total From the table, 20% of patients have a regular heartbeat and low blood pressure.

8 13. Solution: C Let x be the probability of having all three risk factors. [][][]1| B CxPABCABPA Bx = == + It follows that () +=+== Now we want to find ()()[][ ]( ) ( )( )|1 11 3 1 2 APAPA B CPA = = = == Page 6 of 91 14. Solution: A 12301111111 05555555kkkkkp ppppk ==== = 1 = 000000151,4/515415kkkkppppp == ===== Therefore, P[N > 1] = 1 P[N 1] = 1 (4/5 + 4/5 x 1/5) = 1 24/25 = 1/25 =.

9 15. Solution: C Let x be the probability of choosing A and B, but not C, y the probability of choosing A and C, but not B, z the probability of choosing B and C, but not A. We want to find w = 1 (x + y + z). We have x + y = 1/4, x + z = 1/3, y + z = 5/12. Adding these three equations gives ( ) ( ) ( )()()11 543 xz yzxyzxyzwxyz+++++=++++ =++== ++ = = Alternatively the three equations can be solved to give x = 1/12, y = 1/6, z =1/4 again leading to 1 11 1112642w = ++ = 16. Solution: D Let N1 and N2 denote the number of claims during weeks one and two, respectively.

10 Then since they are independent, [][] []712120718079096P7 PPr711 221 2811 2264nnnnnNNNn Nn=+ ==+==== = == = = Page 7 of 91 17. Solution: D Let O = event of operating room charges and E = event of emergency room charges. Then () ( ) ( ) ()( ) ( ) ( ) ( ) () IndependencePO E PO PE PO EPO PE POE= = + =+ Because ( )( ) 1, ( ) EP E PE== =, ( )( )( ) ( )(1 ) ( ) / + = === 18.