CFA Practice Question
Suzanne is researching a cure for a tropical disease and has been struggling to keep her samples uncontaminated. There is a 10% chance that she will forgot to wash her test tube and cross-contaminate the sample, a 18% chance that bacteria will leech into her work, and a 33% chance that her assistant will trip and shatter the sample on the floor. What is the probability that Suzanne's work will not become contaminated?
A. 39%
B. 55%
C. 61%
Explanation: Use the addition rule that one or more of the events will occur, then subtract this from the total probability. P(NOT A) = 1 - [P(A) + P(B)-P(AB)]
User Contributed Comments 11
| User | Comment |
|---|---|
| JeremyMartin | Can anyone provide the solution to this problem? Is the tripping of the assistant related to the contamination? |
| lisab0131 | P(a or b or C) = P(a) + P(B) + P(c) - P(a)*P(b)*P(c) thus P(of none of them) = 1 - P(a or b or c) |
| 7Ricky | Actually P(A or B or C) = P(a)+P(b)+P(c)-P(A)P(B)-P(A)P(C)-P(B)P(C)+P(A)P(B)P(C) 1-P(AorBorC) = Right answer of .49446 another way of getting the same answer is [1-P(A)]x[1-P(B)]x[1-P(C)] = .49446 You can google P(AorBorC) to see the same solution |
| dipu617 | All those probabilities are independent events. That means the total probability of contamination = (10% + 18% + 33%) = 61% So the probability that her work will not be contaminated = (1 - 0.61) = 0.39 or 39% |
| mary11 | I am pretty sure, I will get these questions wrong in the exam. Does not take much to trick me with these questions |
| mary11 | Oh, and I think dipu617 is correct - that was my solution. |
| purich | I think this question is stupid, it doesn't say that they were independent, and it's not an obvious assumption, I think 7Ricky's solution should be correct. |
| jjhigdon | Its a vague question. The "correct" answer assumes the 3 events are independent AND mutually exclusive. 7ricky's solution is correct if you assume the events are independent but not mutually exclusive, which was my initial inclination. But it wasn't one of the answers, so I went with the former... |
| seanj951 | Dipu and mary, I am not sure you can just add the probabilities, though that does give you the "correct" answer here. However, if there were another event with a 39% chance of contaminating then this would lead us to conclude there is a 0% chance of no contamination, which would not be true. I believe Ricky is correct. |
| lynserious | lmao, is it even a financial related problem |
| ctschro | i believe jjhigdon is right about this |