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**CFA Practice Question**

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

A. 1/4

B. 1/8

C. 3/4

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**User Contributed Comments**
22

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

sireklove |
P(Yr 1 >$500m|Yr 2>$500m) = [P (Yr 2 >$500m|Yr 1 >$500m)]/P(Yr 2 >$500m) = [(0.90)(0.10)]/(0.36) P(Yr 2>$500m) = (0.10)(0.90) + (0.90)(0.30) |

sireklove |
oops. Numerator should be [P(Yr 2>$500m|Yr 1 >$500m)P(Yr 1 >$500m)] |

xxxx |
The formula used is Bayes. |

KD101 |
Make it simple: Since we all talk about > 500 mn P(Y1/Y2) = P(Y2/Y1)*P(Y1) -------------------------------- P(Y2/Y1)*P(Y1) + P(Y2/Y1')*P(Y1') Where Y1' = event 1-Y1 This is the standard Conditional Prob formula |

Carol1 |
Use Bay's Formula |

jackwez |
.9 (90% for success rate) x .3 (30% for nonsuccess rate)= .27 which is a good guess in a time crunch |

Birdy101 |
was also making a good guess, but by looking at sirelove formula, i don t get the result of 0.25? can anybody help? |

patccsat |
how do you apply Bayes? I am asking b/c the sum of the conditional probabilities exceed 1. |

patccsat |
got it now... i got confused... need to find the unconditional probability of p(more than 500m in 2nd year) first... which is 36% and go on from there... It's easier to draw the diagram and figure it out (for me)... more intuitive |

scotty21 |
yeah agreed, i just draw the diagram and work it out that way |

Yohan3109 |
very simple, 10% * 90% = 9% to get more than 500. 90% * 30$ to get 500 more when previous was less = 27% So total probability to have 500 more for years 2 is 36% and the probabilité that the first was more than 500 is 9/36 which is 1/4 ore 25% |

mattg |
No formulas required for this one There are 4 possible outcomes: Year 1 Year 2 Outcome 1 Over 500m Over 500m " " 2 Over 500m Under 500m 3 Under 500m Over 500m 4 Under 500m Under 500m Probability for any one of these outcomes is 1/4 |

shiva5555 |
uh, wrong mattg |

Clude |
P(2nd more)=10%*90% + 90%*30%=36% P(1st more/2nd more) * P(2nd more) = P(1st more AND 2nd more) =>X * 26% = 10% * 90% =>X=1/4 |

gshukul |
good stuff Clude...your explanation is the clearest |

mc42086 |
Yes! This one killed me until I realized that. Doing a tree makes it so intuitive. P of >500 >500 = .1 X .9 P of >500 <500 = .1 X (100%-90%)=.01 P of <500 >500 = .9 X .3 = .27 P of <500 <500 = .9 X (100%-30%)= .63 If 2nd year is >500 then 1st and 3rd branch are in play so .9/(.9+.27) = .25 |

Shcote |
(0,1*0,9)/((0,1*0,9)+(0,9*0,3)) |

Nitishm |
Clude example is clearest, except i think it should have been => X*36% = 10% * 90% |

Procbaby1 |
Bayes for a binomial distriubtion |

birdperson |
i used a tree a la mc42086 |

sshetty2 |
the unconditional expected probability in the denominator of the bayes theorem is calculated using mc42's tree from above. (.9/.36)*(.1) |

devleena34 |
It is not understood from Clude's working why 0.1x0.9(2nd more ) considered on the RHS of the equation when 2nd more has been calculated as 0.36 |