[[Exam P-1 (SOA)]] / [[Exam P-1 (SOA) Study Hub|Study]] / ==Quiz==
```quiz
### Viewing habits (3 sports)
---
id: Q1
points: 1
difficulty: medium
---
A survey of a group’s viewing habits over the last year revealed the following information:
(i) 28% watched gymnastics
(ii) 29% watched baseball
(iii) 19% watched soccer
(iv) 14% watched gymnastics and baseball
(v) 12% watched baseball and soccer
(vi) 10% watched gymnastics and soccer
(vii) 8% watched all three sports.
Calculate the percentage of the group that watched none of the three sports during the last year.
- [ ] 24%
- [ ] 36%
- [ ] 41%
- [x] 52%
- [ ] 60%
> Let \(G,B,S\) denote watching gymnastics, baseball, soccer. \(P(G\cup B\cup S)=\sum P-\sum P(\cap)+P(G\cap B\cap S)=0.28+0.29+0.19-0.14-0.12-0.10+0.08=0.48\). None \(=1-0.48=0.52=52\%\).
### PCP visit: lab + referral
---
id: Q2
points: 1
difficulty: medium
---
The probability that a visit to a primary care physician’s (PCP) office results in neither lab work nor referral to a specialist is 35%. Of those coming to a PCP’s office, 30% are referred to specialists and 40% require lab work.
Calculate the probability that a visit to a PCP’s office results in both lab work and referral to a specialist.
- [x] 0.05
- [ ] 0.12
- [ ] 0.18
- [ ] 0.25
- [ ] 0.35
> Let \(R,L\) be referral and lab. \(P(R\cap L)=P(R)+P(L)-P(R\cup L)=0.30+0.40-\big(1-0.35\big)=0.05.\)
### Set identities to find \(P[A]\)
---
id: Q3
points: 1
difficulty: medium
---
You are given \(P[A\cup B]=0.7\) and \(P[A\cup B'] = 0.9\).
Calculate \(P[A]\).
- [ ] 0.2
- [ ] 0.3
- [ ] 0.4
- [x] 0.6
- [ ] 0.8
> Using \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) and \(P(A\cup B^c)=P(A)+P(B^c)-P(A\cap B^c)\); add to get \(0.7+0.9=2P(A)+1-P(A\cap B\cup A\cap B^c)\Rightarrow 1.6=2P(A)+1\Rightarrow P(A)=0.6.\)
### Two urns, same color
---
id: Q4
points: 1
difficulty: medium
---
An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is 0.44.
Calculate the number of blue balls in the second urn.
- [x] 4
- [ ] 20
- [ ] 24
- [ ] 44
- [ ] 64
> Let \(x\) be blue in urn 2. \(P(\text{same})=\tfrac{4}{10}\tfrac{16}{16+x}+\tfrac{6}{10}\tfrac{x}{16+x}=0.44\). Solve \( \frac{ (4)(16) + (6)x }{10(16+x)}=0.44 \Rightarrow x=4.\)
### Young, female, single
---
id: Q5
points: 1
difficulty: medium
---
An auto insurance company has 10,000 policyholders. Each policyholder is classified as (i) young or old; (ii) male or female; and (iii) married or single.
Of these policyholders, 3000 are young, 4600 are male, and 7000 are married. The policyholders can also be classified as 1320 young males, 3010 married males, and 1400 young married persons. Finally, 600 of the policyholders are young married males.
Calculate the number of the company’s policyholders who are young, female, and single.
- [ ] 280
- [ ] 423
- [ ] 486
- [x] 880
- [ ] 896
> \(N(\text{Young,Female,Single})=N(\text{Young,Female})-N(\text{Young,Female,Married})\)
\(=N(\text{Young})-N(\text{Young,Male})-[N(\text{Young,Married})-N(\text{Young,Married,Male})]\)
\(=3000-1320-(1400-600)=880.\)
### Heart disease given parents w/o history
---
id: Q6
points: 1
difficulty: medium
---
A public health researcher examines the medical records of a group of 937 men who died in 1999 and discovers that 210 of the men died from causes related to heart disease. Moreover, 312 of the 937 men had at least one parent who suffered from heart disease, and, of these 312 men, 102 died from causes related to heart disease.
Calculate the probability that a man randomly selected from this group died of causes related to heart disease, given that neither of his parents suffered from heart disease.
- [ ] 0.115
- [x] 0.173
- [ ] 0.224
- [ ] 0.327
- [ ] 0.514
> \(P(H\mid F^c)=\frac{P(H\cap F^c)}{P(F^c)}=\frac{(210-102)/937}{(937-312)/937}=\frac{108}{625}=0.173.\)
### Renewals: auto & homeowners
---
id: Q7
points: 1
difficulty: medium
---
Among policyholders: \(P(A)=0.65,\ P(H)=0.50,\ P(A\cap H)=0.15\). Renewal probabilities: only auto 0.40; only home 0.60; both (at least one) 0.80.
Using the company’s estimates, calculate the percentage of policyholders that will renew at least one policy next year.
- [ ] 20%
- [ ] 29%
- [ ] 41%
- [x] 53%
- [ ] 70%
> Portion renewing \(=0.4P(A\cap H^c)+0.6P(H\cap A^c)+0.8P(A\cap H)=0.4(0.50)+0.6(0.35)+0.8(0.15)=0.53.\)
### PT vs chiropractor
---
id: Q8
points: 1
difficulty: medium
---
Among a large group of patients: \(P(C\cap T)=0.22\), \(P(C\cup T)^c=0.12\). Also \(P(C)=P(T)+0.14\).
Calculate \(P(T)\).
- [ ] 0.26
- [ ] 0.38
- [ ] 0.40
- [x] 0.48
- [ ] 0.62
> \(P(C\cup T)=0.88=P(C)+P(T)-0.22=(P(T)+0.14)+P(T)-0.22\Rightarrow 2P(T)-0.08=0.88\Rightarrow P(T)=0.48.\)
### Exactly one car & not sports
---
id: Q9
points: 1
difficulty: medium
---
Auto customers: all insure ≥1 car; 70% insure >1 car; 20% insure a sports car; among >1 car customers, 15% insure a sports car.
Calculate the probability a randomly selected customer insures exactly one car and that car is not a sports car.
- [ ] 0.13
- [x] 0.21
- [ ] 0.24
- [ ] 0.25
- [ ] 0.30
> \(P(\text{1 car \& no sports})=1-P(>1)-P(S)+P(>1\cap S)=1-0.70-0.20+0.15(0.70)=0.21.\)
### Collision & disability independence
---
id: Q10
points: 1
difficulty: medium
---
Conclusions: (i) collision twice as likely as disability; (ii) independence; (iii) \(P(C\cap D)=0.15\).
Calculate the probability that an automobile owner purchases neither collision nor disability coverage.
- [ ] 0.18
- [x] 0.33
- [ ] 0.48
- [ ] 0.67
- [ ] 0.82
> Independence: \(P(C)=2P(D)\), \(P(C\cap D)=P(C)P(D)=0.15\Rightarrow 2p\cdot p=0.15\Rightarrow p=0.075,\ P(C)=0.15\). Then \(P(C^c\cap D^c)=(1-0.15)(1-0.075)=0.33.\)
### BP and heartbeat table
---
id: Q11
points: 1
difficulty: medium
---
Patients categorized by blood pressure (high/low/normal) and heartbeat (regular/irregular):
(i) 14% high BP; (ii) 22% low BP; (iii) 15% irregular; (iv) among irregular, 1/3 high BP; (v) among normal BP, 1/8 irregular.
Calculate the portion with a regular heartbeat and low blood pressure.
- [ ] 2%
- [ ] 5%
- [ ] 8%
- [ ] 9%
- [x] 20%
> Filling the table from the conditions yields Regular×Low \(=0.20\). (Computed table: Regular: High 0.09, Low 0.20, Normal 0.56; Irregular: High 0.05, Low 0.02, Normal 0.08.)
### Three health risk factors
---
id: Q12
points: 1
difficulty: medium
---
Risk factors \(A,B,C\). For each single-only factor: prob \(=0.1\). For any exact two: prob \(=0.12\). \(P(C\mid A\cap B)=1/3\).
Calculate \(P(\text{none}\mid A^c)\).
- [ ] 0.280
- [ ] 0.311
- [x] 0.467
- [ ] 0.484
- [ ] 0.700
> From \(P(A\cap B\cap C)=\tfrac13 P(A\cap B)=\tfrac13(0.12+0.06)=0.06\). Sum over categories to get \(P(A)=0.40\), \(P(B\cup C\mid A^c)=0.533\), hence \(P(\text{none}\mid A^c)=0.467.\)
### Claim count recurrence
---
id: Q13
points: 1
difficulty: medium
---
Model number of claims in 3 years with \(p(n+1)=0.2\,p(n)\) for integers \(n\ge0\). Find \(P(N>1)\).
- [x] 0.04
- [ ] 0.16
- [ ] 0.20
- [ ] 0.80
- [ ] 0.96
> Geometric tail with ratio 0.2 gives \(p(0)=4/5, p(1)=4/25\). Then \(P(N\le1)=4/5+4/25=24/25\Rightarrow P(N>1)=1/25=0.04.\)
### Pick exactly two of A,B,C
---
id: Q14
points: 1
difficulty: medium
---
Employees choose exactly two of A,B,C or none. Proportions choosing A,B,C are \(1/4, 1/3, 5/12\), respectively.
Calculate the probability a randomly chosen employee chooses no supplementary coverage.
- [ ] 0
- [ ] 47/144
- [x] 1/2
- [ ] 97/144
- [ ] 7/9
> Let \(x=P(AB),y=P(AC),z=P(BC)\). Then \(x+y=1/4,\ x+z=1/3,\ y+z=5/12\Rightarrow x+y+z=1/2\). With “exactly two or none”, \(P(\text{none})=1-(x+y+z)=1/2.\)
### Two-week claims total = 7
---
id: Q15
points: 1
difficulty: medium
---
Weekly claims: \(P(N=n)=\dfrac{1}{2^{n+1}},\ n=0,1,\dots\), independent week to week.
Calculate the probability that exactly seven claims will be received during a given two-week period.
- [ ] 1/256
- [ ] 1/128
- [ ] 7/512
- [x] 1/64
- [ ] 1/32
> Convolution: \(\sum_{n=0}^7 \frac{1}{2^{n+1}}\frac{1}{2^{(7-n)+1}}=\sum_{n=0}^7 \frac{1}{2^9}=8/512=1/64.\)
### ER vs OR charges (independent)
---
id: Q16
points: 1
difficulty: medium
---
\(P(\text{ER or OR})=0.85\). \(P(\text{no ER})=0.25\). ER and OR independent.
Calculate \(P(\text{OR})\).
- [ ] 0.10
- [ ] 0.20
- [ ] 0.25
- [x] 0.40
- [ ] 0.80
> Let \(e=P(E)=0.75\), \(o=P(O)\). Independence: \(0.85=P(E)+P(O)-P(E)P(O)=0.75+o-0.75o\Rightarrow o=0.40.\)
### Averaging two instruments
---
id: Q17
points: 1
difficulty: medium
---
Errors: less accurate \(\sim N(0,(0.0056h)^2)\); more accurate \(\sim N(0,(0.0044h)^2)\), independent. Find \(P(\text{average within }0.005h)\).
- [ ] 0.38
- [ ] 0.47
- [ ] 0.68
- [x] 0.84
- [ ] 0.90
> Mean of two errors has sd \( \sigma=\sqrt{(0.0056^2+0.0044^2)/4}\,h=0.00356h\). \(P(|Y|\le0.005h)=P(|Z|\le 1.4)=2\Phi(1.4)-1\approx0.84.\)
### Accident given accident occurred (age)
---
id: Q18
points: 1
difficulty: medium
---
Age groups with accident probabilities and portfolio shares:
| Age | \(P(\text{acc})\) | Share |
|---|---|---|
| 16–20 | 0.06 | 0.08 |
| 21–30 | 0.03 | 0.15 |
| 31–65 | 0.02 | 0.49 |
| 66–99 | 0.04 | 0.28 |
A randomly selected insured driver has an accident. Calculate \(P(\text{16–20} \mid \text{accident})\).
- [ ] 0.13
- [x] 0.16
- [ ] 0.19
- [ ] 0.23
- [ ] 0.40
> By Bayes: \(\dfrac{0.06\cdot0.08}{0.06\cdot0.08+0.03\cdot0.15+0.02\cdot0.49+0.04\cdot0.28}=0.16.\)
### Death: posterior ultra-preferred
---
id: Q19
points: 1
difficulty: medium
---
Policyholder classes: standard 50% with death prob 0.010; preferred 40% with 0.005; ultra-preferred 10% with 0.001. A policyholder dies in the next year.
Calculate \(P(\text{ultra-preferred}\mid \text{death})\).
- [ ] 0.0001
- [ ] 0.0010
- [ ] 0.0071
- [x] 0.0141
- [ ] 0.2817
> \(\dfrac{0.001\cdot0.10}{0.010\cdot0.50+0.005\cdot0.40+0.001\cdot0.10}=0.0141.\)
### Survived → serious?
---
id: Q20
points: 1
difficulty: medium
---
ER triage: 10% critical, 30% serious, rest stable; death rates: critical 40%, serious 10%, stable 1%. Given that a patient survived, calculate the probability the patient was categorized as serious upon arrival.
- [ ] 0.06
- [x] 0.29
- [ ] 0.30
- [ ] 0.39
- [ ] 0.64
> \(P(\text{Survive})=0.6(0.10)+0.9(0.30)+0.99(0.60)=0.87.\)
\(P(\text{Serious} \mid \text{Survive})=\dfrac{0.9\cdot0.30}{0.87}=0.29.\)
```