The number of days a particular stock increases in a given five-day period is uniformly distributed between zero and five inclusive. In a given five-day trading week, what is the probability that the stock will increase exactly three days? A) 0.333. B) 0.167. C) 0.600. Firm A can fall short, meet, or exceed its earnings forecast. Each of these events is equally likely. Whether firm A increases its dividend will depend upon these outcomes. Respectively, the probabilities of a dividend increase conditional on the firm falling short, meeting or exceeding the forecast are 20%, 30%, and 50%. The unconditional probability of a dividend increase is: A) 0.333. B) 0.500. C) 1.000.
B A
B (1/6)=.167 C On the second ?, they are asking for the UNCONDITIONAL prob of an increase. If the probability of a div increase is unconditional of an earnings beat or miss, it looks like its a sure thing…prob = 1. Am I wrong here?
Hi June 2009, I worked it at 0.333 using a probability tree.
B A Question 1 is a trick, I tried to calculate out the binomial amount, but then clued in and saw the word uniform. For Question two, the unconditional probability can be found by using the total probability rule- so (.333 * .2) + (.333 * .3) + (.333 * .5) = .3333.
soddy and sbmarti - Aren’t your calcs for ?2 the CONDITIONAL prob? I thought ? 2 was a trick question that would make you think the prob tree is the correct way to answer, but since the question asks for UNconditional, i could sum the probabilities.
unconditional means that you find expected value through each condition. Because this value has been routed through each condition (falling short, meeting or exceeding) it is therefore unconditional. I hate probabilities 
Ughhh. thanks.
Seriously though, the great thing about probabilities is that although it is such a hard subject to get a grasp of, once you cop on to it, it just all seems to fall nicely into place. Trees and diagrams all the way…
B C for second question P(short earnings)=1/3 P(meeting earnings)=1/3 P(exceeding earnings) =1/3 (coz question says they all are equally likely) so, P(Div inc)= P(Div inc/short)* P(short) +P(Div inc/meet)*P(meet) + P(Div inc/exceed)*P(exceed) = .2 * 1/3 + .3 *1/3 + .5*1/3
Gauri, you wrote C for the 2nd question but your math comes out with answer A?
ohh yeah … its A got the numbering backwards…usual exam mistake!
I got #2 no problem but missed #1. Can someone please explain? Thanks, MrE
1/6
Discrete uniform distribution: P(X=x)=1/N (where x=x1,x2, … xN , remember total number of values is N); so in question1 P(X=x) =1/6 where x=0,1,2,3,4,5 total 6 values where each value of x represents the number of days a stock can increase in a give 5 day period and therefore P(X=3)=1/6 MrE2All Wrote: ------------------------------------------------------- > I got #2 no problem but missed #1. Can someone > please explain? > > Thanks, > MrE