- CFA Exams
- CFA Level I Exam
- Study Session 2. Quantitative Methods (1)
- Reading 8. Probability Concepts
- Subject 5. The Total Probability Rule

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**CFA Practice Question**

A study found that, on average, 10% of a pharmaceutical company's drugs that are placed on the market make more than $500 million in their first year. If a drug makes more than $500 million in its first year on the market, its probability of making more than $500 million in the second year goes up to 90%. If, on the other hand, the drug makes under $500 million during its first year on the market, its probability of making more than $500 million the second year is only 30%. If a drug made $750 million in the second year, what is the probability that it made more than $500 million in the first year after its launch?

A. 1/4

B. 1/8

C. 3/4

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**User Contributed Comments**
16

User |
Comment |
---|---|

Eckhardt1978 |
i do not understand this question, and how i can reach this answer. |

tenny45 |
I got it correct by guessing but I don't really know the calculation to this question either. |

CocaColas |
Could it be that there was 1/2 chance of reaching $750m in the second year and having reached $750m in the second year, the drugs' sales in the first year could either have been less than $500m or greater than $500m, hence 1/2*1/2=1/4? |

antarctica |
I got 36% and could not choose anything but the closest figure and got the answer correctly...by chance? |

melmilesxx |
This is total probability and Bayes Theorum. First find the total probabity of earning over 500 in the second year (.90*.10) + (.30*.90) = .36. Using Bayes Theorum, the prob of earning over in year 1 givin over in year 2 = P(over in year 2|over in year1)*P(over in year one)/ P(over in year 2) = (.90*.10)/.36 = .25 |

Shelton |
p(5P_1|5P_2) = p(5P_2|5P_1)*p(5P_1)/ [p(5P_2|5P_1)*p(5P_1)+ p(5P_2|5M_1)*p(5M_1)] = (0.9*0.1)/(0.9*0.1+0.3*0.9) = 1/4 |

egghead |
The same in words: it could be above 500 for the 2dn year in two ways: 1) bellow 500 1st year and above 2nd year: P=0.9*0.3=0.27 2) above 500 1st year and above 2nd year P=0.1*0.9=0.09. Totally its 0.36 and path 2) contribute 0.09 or 1/4 of it, |

cfandidate |
to eggplant: thanks, your explanation was the most understandable! |

panvino |
Use the tree diagram!!!!!!! |

Adzz |
tricky question, thx egghead for the explantion! |

Gooner7 |
Listen to Panvino use the tree! |

Mikehuynh |
Would be easier with a diagram, but not enough time during the exam. Try my luck |

CJPerugini |
Easiest Method: Pretend there are 100 different drugs and draw a tree diagram as if you didn't know the drug made $750M the second year. You should have: 10 drugs made over $500M the 1st year and of those, 9 made over $500M the second year and 1 did not. 90 drugs made less that $500M the 1st year and of those, 27 made over $500M the second year and 63 did not. Next eliminate the companies that the additional information does not pertain to. You are left with: 9 companies that made over $500M the first year 27 companies that made less than $500M the first year Probability the company made over $500M the first year? 9/(9+27) |

Murtadha |
Letters are being used for visual purposes: A - Y1 above 500 = 0.1 B - Y2 above 500 = 0.09 C - Y2 below 500 = 0.1 D - Y2 less than 500 = 0.9 E - Y2 above 500 = 0.3 F - Y2 below 500 = 0.7 A x B = 0.09 A x C = 0.01 D x E = 0.27 D x F = 0.63 Make sure all probabilities add up to 1 as a sanity check. Answer = A x B / { (A x B ) + (D x E) } Answer = 0.09 / (0.09 + 0.27) = 0.25 |

ascruggs92 |
Simply put, 9/10 companies that make 500M in Y1 will make more in Y2, and only 3/10 that do not make $500M in Y1 will make over that in Y2. If they were actually 10 of each of these companies, 12/20 will make $500M the next year. Of those 12, three will have not made $500M the year prior. 3/12 = 25% |

JClune |
P(B)=prob of more than 500 in 2nd year P(A)=prob of more than 500 in 1st year. We want to know P(A|B) which is =P(AB)/P(B) Using a tree diagram P(B)=0.1x.9+.9*.3=0.36 P(AB)=0.01*0.09=0.09 So P(A|B)=0.09/0.36=0.25=1/4 |