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PROBABILITY PROBLEMS (AND SOLUTIONS)  Q.1-21 from the notes

        Some additional practice problems in probability have been added at the end of this section ..............

 

 

All questions have appeared as questions in the mid-term or the final exam at some time.  Numbers 20 and 21 are a little more challenging, and while they are not required at undergraduate level, you are invited to try them as they may help refine your skills and understanding.   

Part 1: Solutions to Q. 1-21 in the course notes (note: some have links to detailed notes and diagrams).
1.   (a) 0.333   (b) 0.5    (c) 0.25
2.   (a) 0.8464   (b) 0.0064    (c) 0.1472  
3.   (a) 0.025   (b) 0.55   (c) 0.6818
4.   (a) 0.900   (b) 0.100)
5.   (a) 0.40   (b) 0.60    (c) 0.70   (d) yes
6.  (a) 5.988x10-3  (b) 0.994012
7.    (a) 0.79515   (b) 3.5x10-4    (c) 0.01445   (d) 0.99965
8.  (a) 272   (b) 0.294   (c) 0.5   (d) 0.34 (note that this assumes 'not both'.  If with both = 0.44)
9.  (a) 0.875   (b) 0.125   First decision is 70% vs 30% of fuses: 0.7 for X  vs 0.3 for Z .  Next juncture, failure rates for each X: 0.7 v 0.3  and  Z: 0.1 and 0.9   
10.  (a) 0.48  (b) 0.44   (c) 0.08
11.   Dialysis:68% survival v. transpl: 66% survival
12.   (0.8)20 = 0.01153
13.   (a) .00498    (b) 0.99976
14.   (a) 0.11385    (b) 0.99967    (c) 0.88615
15. .(a) 0.099625   (b) 3.75305x10-4   (c) 0.99625   (d) 0.375%      Solution:  use days/yr for your probabilities.   Hence the first  FSD failure is 20/365 and non-failure is 345/365, 2nd FSD failure is 25/365, and so on.  [A] and [B] calc from the origin (annual need), while [C] assumes already the need to shut down, so take the calc downstream from the 0.1 annual need. [D] reduce the tree to per-year prob of serious incident (0.000187653).  Take the complement (0.999812347) as the prob of NO serious incident in one yr.  Raise this to power of 20 (20 years), or 0.9962536, and take the complement again to get the prob of any incident in 20 years (0.003746, or 0.375%)    DIAGRAM IS SHOWN HERE
  16.     (a) 1.62%  (b) 0.108  (c) 0.8878  (d) 706 more survived due to partial Tr than if no Tr avail;  (e) 176 more d could have been prev by 100% Tr     NOTE :  The key approach to (d) is to compare survivors now (with partial tr) =1605.6 against the expected number of survivors if NO tr was available : in other words at a 50% fatality rate: (1800)(0.50)=900.  Difference is 705.6   Similarly, [e]  compares deaths now (with partial treatment) = 194.4  with deaths if tr were available for all - in other words a 1% fatality rate: (1800)(0.01)=18 leaving a difference of = 176
17.  (a) 0.99  (b) 8x10-3  (c) 7.912x10-3  (d) 8.784x10-5  (e) 8x10-8   (f) 1x10-5  (g) 0.99999   (h) 2x10-3
18.  (a) 0.025   (b) 0.999   (c) 1.00139x10-5   (d) 0.28066  (e) 80 (8 per yr)      NDOTES:   The tree itself is the tricky part here, especially the insertion of the given probability 1/100,000.   Look carefully at the wording.  It is NOT a conditional probability.  It is a JOINT probability, and therefore cannot be at an intersection.  It has to be the product of Virus+ and Test   (therefore a joint probability), and if you look at the false positive value (1/1000) you will see that is sounds like a conditional probability and therefore goes at the intersection for T+ or T given that the Virus is +.     You will need to work backward to find the conditional probability Test given that V+    You do this of course by dividing 0.00001 by the incidence rate  (4 in 10,000), to yield 0.025.    The rest of the table is familiar.  DIAGRAM IS SHOWN HERE
19.  (a) 2.4x10-3        (b) 0.1184         (c) from 0.1184 to .002517, a drop of 0.115879 or 97.9%     (d) [note: calculated as 25 year period, not 10 years] liability is reduced from $947,192 to $20,160 (97.9%)
20. (a) 3.74953x10-4      (b) 7.496188x10-4       (c) 1.4058975x10-7     (d) (0.99925038924 = 0.5 thus 924/10 - 92.4 weeks = 1 yr 40 weeks) Approx 1 April 1998    Explanatory notes   (a) First take the risk for 1 hour (= 1/8000) and find the risk for 3 hrs.   Multiplying by 3 is very close, but the correct way is to take the risk for NOT failing in 1 hr (7999/8000) and raise to the power 3  Then take the compliment again.   (b) New tree:  first engine fails or not, and then 2nd engine fails or not.   Four outcomes:  add the two middle ones = exactly one engine fails.    (c) This is the top joint prob (both fail)   (d)  This has to be worked backwards.   The risk of failure on one engine in a 3 hr flight is calc. as in (b).  The JOINT probability is given as 0.500000.       So you need the exponent for the complement for (b) that will reach 0.500000    Recall that when  Xn = Y,   then  n = logY / logX     This produces about 924 flights.   Work back to calculate the date.
21.

(a) 0.16279    (b) 28.6% more likely     (c) 83.7    (d) 0.05      (e) 60     (f) $150,000      

 Explanatory notes:       

(a) 0.0035/(0.0035+0.018) = 0.16279         

(b) P(def|maint)= 0.35;   P(def| n/maint)= 0.45;  difference=0.1, expressed as increase over base level: 0.01/0.35 = 0.2857  

        (Note that as P(acc|def) = 0.05 for both levels of maintenance, this can be ignored, with (Pdef) being used directly.        

(c) (incorrectly maint + def)/(all defective) = (0.36)/(0.36+0.07) = 0.837

(d) 0.05 (given)        

(e) existing 20% maintained of 60,000 trucks produces (0.0035+0.0180)(60.000) = 1,290 accidents; hypothetically, 40% maintained of 60,000 trucks would

        yield (0.007+0.0135)(60,000) = 1,230 accidents, or 60 fewer.      

(f) if 20% additional trucks properly maintained results in 60 fewer accidents, then 1% increase in maintenance (costing $450,000) should

      avoid 3 accidents.   So, prevention of one accident should cost $150,000

   
   

 

Supplemental PRACTICE PROBLEMS (NOT IN THE COURSE NOTES)    The answers are also included (scroll down)

Suppl. 31A.   [This is the original form of this question.  Pay special attention to the exact wording. The solution is shown below]
Assume the following: 1 in every 150 children are allergic to peanuts, and  1 in every 250 children are allergic to milk.   These allergies are completely independent.   What is the probability of selecting one child, at random, who is:   (i) - allergic to BOTH milk and peanuts?        (ii) - allergic to ONLY peanuts?        (iii) - allergic to peanuts OR milk (including both)?     (iv) - allergic to NEITHER peanuts NOR milk?     (v) - What is the probability that a class of 47 children will contain a child who is allergic to AT LEAST ONE of the two foods?

 

Suppl. 31B   [This is a Variant of 1A .]  Again, read the question carefully.
Assume the following: 1 in every 150 children are allergic to nuts, and  1 in every 250 children are allergic to tartrazine food colour.  We do not know if these allergies are independent.   We DO know that 1 in 350 children have BOTH allergies.  What is the probability of selecting one child, at random, who is:           (i) - allergic ONLY to nuts?        (ii) - allergic to nuts OR tartrazine (or both)?     (iii) - allergic to NEITHER nuts NOR tartrazine?     (iv) - What is the probability that a class of 50 children will contain a child who is allergic to AT LEAST ONE of the two foods?  And  (v) - are the two allergies independent?

Supp. 32.

HIV

TB pos.

 

TB  neg.

Totals

      A survey of 5,260 people in a developing country reveals that approximately 1 in 120 is serologically positive for TB  and 1 in 90 is positive for  HIV. The actual results  are  shown at left. (Hint: enter the (overall) probabilities in parentheses.)

pos.

3

 

55

58

neg.

41

5161

 

5202

44

5216

5260

 

      (i) -   are the two infections independent?
     (ii) -  what is the probability of a person being positive for both infections?
    (iii) -  given that a person is positive for TB, what is the probability that they are also positive for HIV?
    (iv) -  is it more likely that a TB+ve person would also be HIV+ve, OR that a HIV+ve person would also be TB+ve?

 

 

Supp.  33    The potential for dental high-speed drill equipment to spread infection has been recently examined. Particular interest has focussed on the ability of very small quantities of tissue (and any infectious agent present) to remain in the hand-piece and allow for transmission of infection between patients. HIV has been the main concern to date. For illustration purposes, assume that some tissue remains in the hand-piece after 22% of procedures (assume one procedure per patient for this example). Also assume that one in 2,000 patients are HIV +ve. For actual transmission to occur, the following patient would need to have a procedure involving direct tissue damage. Assume that transmission would occur in 1 of 3 patients.    (i) What is the risk of HIV transmission for a random patient at this dental practice?    (ii) What is the risk that one or more transmissions of HIV would have taken place after 100,000 patients?

 

Suppl.  34.       When the first 10 cases of vCJD were publicly recorded in March 1996, it was quickly observed that all were of the same genetic type: homozygous at codon 129 for methionine. Only 40% of the UK population fall into this category, and at the time, it was not known whether this was a coincidence or a characteristic of this prion disease. Calculate the probability of this situation occurring by chance alone; in other words, in the absence of any real mechanism.

 

Suppl.  35.      An aircraft is held together with thousands of aluminium-alloy rivets. After the rivet is placed, it is tested by a portable X-ray device to examine the strength of the join. The x-ray process used for detecting faulty riveting in aircraft construction will correctly detect all but 1 in a thousand faulty rivets. It also wrongly identifies 5 in a thousand perfectly sound rivets as "faulty". The actual rate of faulty rivet operations is 15 in 1,000.   (A) What is the probability that a rivet which shows "OK" on x-ray, is in fact faulty? (B) What percentage of rivets are faulty and erroneously found OK by the x-ray scan? (C) If the completed aircraft contains 420,930 aluminium rivets, (assume that only those that passed X-ray test still remain) how many would be expected to have escaped detection and hence pose a threat to the safety of the aircraft? (D) Would it be better (safer) to improve the X-ray equipment such that it would only fail to detect 1 in 5,000 rivets, OR to improve the initial riveting process such that only 5 would be faulty in each 1,000? Limited resources allow only ONE change.

  

Suppl. 36.    The table below shows results from a survey of children in a favela near So Paulo. The incidence of intestinal worms, and the incidence of protein deficiency syndrome are recorded. (A) what is the probability of finding a child (at random) with both conditions? (B) What is the probability of finding a child with neither condition? (C) what is the probability of finding a child with protein deficiency, given that parasites are present? (D) Given that a child has protein deficiency, what is the probability that it also has parasites? (E) Are the two conditions independent?

                                                          Data for Q 36: Intestinal parasites?
  Pos Neg Tot

Protein                     Yes deficiency?                  No

23

97

  28

322

  51

419

                                        Tot 120 350 470

 

Suppl. 37.    There are 6,200 records of rear-end collisions at speeds >35 K/H involving a certain model of car. It is found that in rear-end collisions, the fatality rate overall is 11.84%. In seven percent of these rear-end collisions the gas tank ruptures, leading to a fire, and in this sub-group, the fatality rate is 23%.

(A) What is the probability of surviving a rear end collision in which a fire developed? (B) What is the probability of surviving a rear-end collision in which a fire did NOT develop? (C) What proportion of fatalities in these rear-end collisions are due to the fire? (D) If the design had prevented the gas tank fire in all but 1 percent of collisions, how many lives would have been saved? (E) If the cost of this better design was $450,000, what will be the (approx) cost per life saved?

                

Suppl. 38   

Extreme chemical sensitivity is a life threatening situation which can be fatal in a few minutes. When the P35 gene appears in a mutated form, (we can call the condition Px) there is no apparent effect on the individual's sensitivity to environmental chemicals.  But if the individual also has a defective enzyme (we can call this condition Ex), then the individual is at a 50% risk of developing a life-threatening sensitivity to certain plasticizers in common use in industry and the environment. (Separately, these factors do not increase a person's risk at all).  One person in every 340 has the mutant gene, and fifteen percent of the population has the defective enzyme.  We do not yet know if the defective enzyme is dependent upon or related to the mutated gene.  A recent survey shows that 53 in 120 000 persons are carrying the mutated gene AND also the defective enzyme.   Construct a tree with these data and answer the following questions:
(A)

 

 (A)   What proportion of the population is free of BOTH factors?

 (B)   What proportion of the population carries the mutated gene but NOT the defective enzyme

 (C)   Are the mutated gene and the defective enzyme independent or dependent factors?

 (D)   Among the individuals with the mutated gene, what is the risk of having the defective enzyme?

 (E)   Among the individuals with the defective enzyme, what is the risk of having this life threatening chemical sensitivity?

 (F)    Among individuals with the mutated gene, what is the risk of having this life-threatening chemical sensitivity?

 (G)   in this population of 120 000 persons, how many would be at risk of such a sensitivity? 

 

Suppl. 39  

 

            One percent of young children in Canada are allergic to peanuts.  A janitor drops a cookie from his lunch box in a preschool room by accident. It contains peanuts.
 

A.     Two preschool kids find the cookie and decide to share it.  What is the probability that at least one of them is allergic to peanuts?

 

B.     What is the probability that in a kindergarten class of 20 kids, at least one would have a peanut allergy

   Roughly 3.0 percent of children are allergic to seafood.  Considering the co-existence of both of these allergies, please calculate the following probabilities. (A table of probabilities for these data is given below)

 

C.C.  P(allergic to seafood given that they are allergic to peanuts)  [0.3]

D.D  P(allergic to peanuts given that they are allergic to seafood)  [0.1]

E.E   P(allergic to both peanuts and seafood) [0.003]

F.FF.  P(being allergic to at least one of these allergens (peanuts, seafood) [0.037]

G.G   Are these allergies independent of each other? [No-  Dependent!]

 
 
seafood allergy
no seafood allergy
tot
peanut allergy
0.00300
 
0.00700
0.01000
No peanut allergy
0.02700
 
0.96300
0.99000
 
tot
 
0.03000
 
0.97000
 
1.00000

 

SOLUTIONS ..... to Supplemental questions above...
Suppl.31A:     (i) 0.00002667    (ii) 0.00664033     (iii) 0.01064033     (iv) 0.9893570       (v) 0.395227

Suppl.31B:     (i) 0.0038096     (ii) 0.0077956     (iii) 0.9922044       (iv) 0.323838   (v)  no - they are dependent   (for tree diagram click here)

Suppl.32.       (i) no    (ii) 0.00057    (iii) 0.06818   (iv) Yes.  (0.06818>0.05172)

Suppl.33.       (i)  0.00003667   (ii)   0.97444103

Suppl.34.      (0.4 raised to the power of 10) = 0.000105

Suppl.35.    A: 1.5305 x 10-5        B: 1.5000 x 10-5          C: 6.3         D: If detection improved to 1/5000, false "OK" rivets would reduce to 3E-06,  but if initial riveting was improved to 5/1000 faulty, the false "OK" rivets would reduce to 5E-06.  Thus the first choice is safer. (note this answer supersedes the previous one)

Suppl.36.   A: 0.0489 B: 0.6851 C: 0.1917 D: 0.4510 E no

Suppl.37. (a) is read from chart as 0.77

               (b) is read from chart as 0.89

               (c) is arguable... non-fire collisions yield 11% fatality, but fire collisions yield 23% fatalities.  Some of these would still be fatal even if no fire present.  So the better answer might be to take the difference (12% and assign this as the proportion of fatalities due to the fire)

              (d) fatality count on 6200 collisions is (6200)(0.0161+0.1023)=734.08 (as shown).  If fire incidence reduced to 1% from 7%, the fatality probabilities are 0.0023 and 0.1089.  Thus on 6200 collisions, the number of fatalities should be (6200)(0.0023+0.1089)=689.44.  The difference is about 45 lives saved at a cost of 450,000, or about 10,000 per life.

 

                                                                                  yes:0.23      ------------->[ 0.0161 ] fatalities 

                                 Yes: 0.07   -------------  fatal?  <     

                                                                               no: 0.77     -------------->[ 0.0539                

   -------------- Fire?     

                          \                                                   yes: 0.11   --------------->[ 0.1023 ] fatalities 

                                 No: 0.93 ---------------  fatal?  <

                                                                                  no :0.89      ------------->[ 0.8277 ]                 

 

 

Suppl. 38 solution  NEW  NOTE: The full TREE is supplied but on some platforms the graphics are not seen.  I am trying to fix the problem.

In the meantime, if you need the view of the tree, send me an email.

The process here is NOT to put in the second level conditional probabilities (risk of def. enzyme = 0.15)

because we are told that we do NOT know if the enzyme deficiency is dependent upon the defective gene.

So the approach is to insert the top joint probability (53 in 120 000 persons are carrying the mutated

gene AND also the defective enzyme), which is 0.00044167

 

The first and third joint prob from the top must add to 0.15, so we can find 0.1495583

sensitivity: .0002208

 

 

Text Box: sensitivity: .0002208
 
 

Then we can calculate 0.15016660 from .00044167 divided by 0.0029412.   Now we can see that

these two variables are INDEPENDENT because we have almost perfect 0.15 and 0.85.  Hence,

the other branches must also be 0.15 and 0.85.
0.5

 

 

 

 

 

 

 

 

 

 

 

 

.00249953
Text Box: .00249953

0.15016660

.8475005
Text Box: .8475005

Go:   0.9970588

Gx:   0.0029412

 

.0002208
Text Box: .0002208

 

0.8500005

0.8498334

0.1499995

 (A)       [0
0.5
.8475]

 (B)       [0.002499]

 (C)       [INDEP]

 (D)       [0.150]

 (E)       [0.001472]

 (F)       [0.0750833] 

 (G)       [26 or 27]

 

 

 Suppl. 39 solution   NEW

A.   [0.0199]

 B.  [18.2%]

C.   [0.3]

D.  [0.1]

E.E    [0.003]

F.FF  [0.037]

G.G    [No- they are Dependent!]

 

 

SOLUTION TO EXAMPLE 4 (PAGE 8) PROBLEM SET

Q4A Q4B Q4C Q4D Q4E Q4F Q4G Q4H Q4 I Q4J
3.80E-06 2.00E-07 4.00E-06 1.00E-05 9.76E-06 2.495E-05 2.0942E-04 1.50E-03 5.0E-03 2.5E-03

(this is half the risk of helicopter pilots)

 

SOLUTION TO EXAMPLE 23  (PAGE 20) PROBLEM SET (Apologies for confusing answers. 

                               ....The handbook was changed slightly after these answers were added to the site.)

Here are some solutions to the question set number 23 (page 20)

23a   P(non-path AND non-res)  +   0.8163   (Note that "neither/nor"  =  "not A AND not B"

23b   P(path)     =  0.1020408

23c    P(Res)      = 30/245 = 0.122

23d    P(Path and Res)  = P(R) x P(P|R) = (30/245) x (10/30) = (10/245) = 4.081633E-02

23e    0.1836735

23f    0.33333

23g   P(Res|Path)  =   10/25  = 0.400

23h   0.0612245

23i   (REPEATED FROM 23e)

23j   0.8979592

 
 

 

 

      

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