RISK ASSESSMENT AND COMMUNICATION
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.
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?
(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 São 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?
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?
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) 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?
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!]
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 ]
\ 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
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.
(G)[26 or 27]
Suppl. 39 solutionNEW
G.G [No- they are Dependent!]
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.)