### CFA Practice Question

There are 410 practice questions for this study session.

### 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

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
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!!!!!!!
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