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Koru81
Jan 29, 2008, 11:50 AM
I'm having some trouble understanding how to set up the following questions. Help please
Billy has submitted applications for two separate research grants, A and B. He feels the probability of getting grant A is 0.50, while the probability of getting grant B is 0.40.
The probability he will get at least one grant is 0.75.
A) Probability of receiving both grants
B) Billy gets grant B, what is the probability of getting grant A?
C) What is the probability of getting A or B, but not both?

For A I chose 0.50 + 0.40 = 0.90 or 90%
I wasn't sure how to attempt B
For C I chose P(A) = 0.50 P(B) = 0.40
P(A) + P(B) - P(A & B)
= 0.50 + 0.40 - 0.75
= 0.15 = 15%

Thanks in advance for some help

jiten55
Jan 29, 2008, 04:40 PM
Deleted because of wrong answer.

See below for the correct answer.

PolluxCastor
Jan 29, 2008, 05:48 PM
The probabilities are based on Billy's feelings, it might be best not to assume the events are independent.

Jiten is correct on A.

B and C present a problem. If the two events are independent (as they should be), the probabilities are inconsistent (again, as Jiten stated).

Might be best to request clarification, Are the probabilities of receiving the grants intended to be independent?

Koru81
Jan 29, 2008, 06:00 PM
Well, I was assuming that they were not mutually exclusive. Actually, part D is are they or are they not mutually exclusive events?

PolluxCastor
Jan 29, 2008, 07:14 PM
Independent and mutually exclusive are not the same.

An example of mutually exclusive is - the probability it is raining - the probability the sky is clear.

Independence is where one event has no effect on the other - tossing a fair die or coin.

I would be inclined to believe the acceptance of a research grant would not be independent of a second application. If one group laughed the grant request out of town, it would be likely a second would also.

I'm going to have to think on the implications of the events not being independent (if indeed that is the case).

What might be helpful in this case is a Venn Diagram - Venn diagram - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Venn_diagram)

Take a look there and see if that helps.

jiten55
Jan 29, 2008, 07:15 PM
"The probabilities are based on Billy's feelings, it might be best not to assume the events are independent."

In case of "feelings", Billy has to make sure that his "feelings" are not inconsistent, which often is the case with "feelings" - that's why feelings are so unreliable!

I look forward to how PolluxCastor handles the possibility of Dependence.

jiten55
Jan 30, 2008, 01:19 AM
AT LAST!

Using VENN diagrams, I get the following results (No presumption of Independence!)

P(AUB) = P(A) + P(B) - P(A INTERSECTION B) = .75 (A or B: At least one grant)

Hence .75 = .5 + .4 - P(A AND B) [Intersection means both!]

P (A AND B) = .15 (Prob of both grants)

P(A|B) = P(A intersect B)/P(B) = .15/.40 = .375 (Prob of A, Given B)

P(A or B but not both) = .75 - .15 = .60

Koru81
Jan 30, 2008, 08:17 AM
Wow, thanks for the help. I'm still meeting with my professor today. I'm not entirely sure I still feel confident enough to answer a similar question.

jiten55
Jan 30, 2008, 08:26 AM
I am confident you will have no problem.

Remember, however, the following probability theorem which I used:

P(A|B) = P(A intersection B)/P(B)

P(A|B) means Probability of A when B is true.

P(A intersection B) means Probability of (Both A AND B)

[When A and B are sets, it means intersection of A and B]

Apart from this, it is just Venn diagram.

jiten55
Jan 30, 2008, 09:38 AM
See attachment - May clarify further.