Probability problem.

• Sep 4th 2012, 02:21 PM
Romanka
Probability problem.
Please, help me! ( I think my difficulties with it because of my poor English (Headbang) a cause- result? or intersection of 2 events? )
Ann estimates that there is a 60% chance that she will get a raise if she asks her boss.
a) If there is a 40% chance that Ann will, in fact, ask her boss for a raise, what is the probability that she will get a raise because she asked a one?
b) If there is a 10% chance that Ann's boss will give her a raise before she even asks for one, what is the probability that Ann will get a raise without asking for one?

I see that it's about conditional probability, but can't get the answer.
Thank you!
• Sep 4th 2012, 03:01 PM
MaxJasper
Re: Probability problem.
My guess is that

• Sep 4th 2012, 04:42 PM
Romanka
Re: Probability problem.
Why is Prob(raise|not ask)=.10 ?
• Sep 4th 2012, 04:48 PM
MaxJasper
Re: Probability problem.
Part 2 of your original post says it!
• Sep 4th 2012, 05:33 PM
Plato
Re: Probability problem.
Quote:

Originally Posted by Romanka
Ann estimates that there is a 60% chance that she will get a raise if she asks her boss.
a) If there is a 40% chance that Ann will, in fact, ask her boss for a raise, what is the probability that she will get a raise because she asked a one?
b) If there is a 10% chance that Ann's boss will give her a raise before she even asks for one, what is the probability that Ann will get a raise without asking for one?
I see that it's about conditional probability, but can't get the answer.

Frankly I doubt you have received sound replies yet.
We have no way to know that the events are independent!
If they are independent, then conditional probability is a moot question.
Basically, I find this a deeply flawed question. It may be a translation problem.
We need more clarification on what the question actually says.
• Sep 4th 2012, 05:51 PM
SworD
Re: Probability problem.
Well they are dependent by definition, whether or not she asks for the raise affects the probability that she gets one, according to the given values. I do agree that the wording is misleading so you might disagree with this..

The "because she asked for one" can be interpreted as, "what is the probability that she asked for a raise AND that she received one." In this case, it would be P(ask)*P(get raise|ask), where. So it would be 0.4 * 0.6 = 0.24.

The second question is worded confusingly, because it says "before she even asks." This implies that she might be considering asking for a raise but her boss might give her one before she even had the chance to ask.

If we interpret it literally... the answer is simply 10%.. the question is basically a rephrase of the very first statement in that problem.. "there is a 10% chance she will get a raise before she asks for one".. this implies that the chance that she does get a raise without asking for one is indeed simply 10%.. but then, it might be a problem with the wording.
• Sep 4th 2012, 05:53 PM
Romanka
Re: Probability problem.
That is the original question from the test (not my translation). I thought I can't clarify it for myself and understand because of my English, but you are agree with me- it's "a deeply flawed question" :(((
• Sep 4th 2012, 06:01 PM
Romanka
Re: Probability problem.
Quote:

Originally Posted by SworD
it might be a problem with the wording.

the only explanation that I have :(
• Sep 4th 2012, 06:11 PM
SworD
Re: Probability problem.
Well I interpreted the wording literally and gave you answers, is there any way you can verify?
• Sep 4th 2012, 06:20 PM
MaxJasper
Re: Probability problem.
Another interpretation of the problem can result in:
$P(r\cap a)=.40$
$P(r\cap \text{na})=.10$
$P(r|a)=.40/.50=.80$
$P(r|\text{na})=.10/.50=.20$