### CFA Practice Question

There are 410 practice questions for this study session.

### CFA Practice Question

The staff of a company is 45% female. The probability of a female requesting a sick leave is 12% versus 10% for males. An individual who requested a sick leave was randomly selected from the staff of this company. The probability that this individual is female would be closest to ______.
A. 45%
B. 50%
C. 55%
Explanation: Use Bayes' formula to compute this probability. Assume the following events:
F: individual is a female, P(F) = 0.45
M: individual is a male, P(M) = 0.55)
S: individual requests a sick leave, P(S|F) = 0.12 and P(S|M) = 0.10
P(F|S) = [P(S|F) x P(F)] / [P(S|F) x P(F) + P(S|M) x P(M)] = 0.495

User Comment
linr0002 total prob rule:
P(S)= P(S/F)*P(F)+P(S/M)*P(M)=0.109

Bayes rule:
P(F/S)= [P(S/F)*P(F)]/ P(S)=0.495
wollogo In simple terms. The probability that the individual was a female is the probability of a female taking a sick day / probability of either male or female taking a sick day.

P(F/S) = P(FS)/P(S) where
P(FS) = P(S/F)*P(F)
MattyBo linr0002's explanation pulls it together. thanks.
sergashev 45*0.12=5.4
55*0.1=5.5
5.4+5.5=10.9
5.4/10.9=0.495
aakash1108 thanks @linr. Good clarity!
JimM That's not how I read the problem at all. I read it as, "A person who requested sick leave was chosen. What is the probability that the person is female?" The calculation was 12 females out of 22 possible choices, therefore 55%. The 45% female population was a red herring.
scprior agree with jim
atemple315 Me too - I read it twice just to be sure and thought I'd done well to spot the "trick" wording.
azramirza 0.45*0.12/0.45*0.12+.55*.10)
PeterL Yes totally agree with Jim
Shaan23 No guys. it says an individual requested sick leave. <---- That is the given. What is the probability its a female?

P(F/Sick)
praj24 F(45) x .12 = 5.4
M(55) x .1 = 5.5

5.4/10.9 = 49.5% >> 50%