CFA Practice Question

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

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
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