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Probability DS

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Probability DS

by aj5105 » Tue Jan 13, 2009 7:07 am
If the probability that event A will occur is 0.5, what is the probability that event B will occur?
(1) The probability that both A and B will occur is 0.3.
(2) The probability that at least one of the events will occur is 0.8.


P(A or B) = P(A) + P(B) - P(A and B)

1)P(A and B) = 0.3 --Not sufficient

2)P(A or B) = 0.8 --Not sufficient

1) + 2) Sufficient

Is my solution right? (especially St 2)
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by welcome » Tue Jan 13, 2009 7:25 am
IMO - B.

Here is the way I proceeded..

for using P(A or B) = P(A) + P(B) - P(A and B)
A) is NOT SUFF
b)
The probability that at least one of the events = 1 - Prob of no event (P(NO)).
0.8 = 1 - p(NO)
P(NO) = 0.2

P(NO) = 0.2 = (Prob of NOT A). (Probs of NOT B) = (1-0.5)(1-p(b))
=> 1-p(b) = 0.2/0.5 => p(b) = 3/5

Ans B.
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by logitech » Tue Jan 13, 2009 9:06 am
welcome wrote:IMO - B.

Here is the way I proceeded..

for using P(A or B) = P(A) + P(B) - P(A and B)
A) is NOT SUFF
b)
The probability that at least one of the events = 1 - Prob of no event (P(NO)).
0.8 = 1 - p(NO)
P(NO) = 0.2

P(NO) = 0.2 = (Prob of NOT A). (Probs of NOT B) = (1-0.5)(1-p(b))
=> 1-p(b) = 0.2/0.5 => p(b) = 3/5

Ans B.
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by yalanand » Tue Jan 13, 2009 12:00 pm
beautifully explained
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by 720dreaming » Tue Jan 13, 2009 12:09 pm
How is A not sufficient?

(.5)(prob B)=(.3)

Prob B =60%

Am I missing something?
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by logitech » Tue Jan 13, 2009 4:18 pm
720dreaming wrote:How is A not sufficient?

(.5)(prob B)=(.3)

Prob B =60%

Am I missing something?
P(A or B) = P(A) + P(B) - P(A and B)
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by 720dreaming » Tue Jan 13, 2009 6:20 pm
Thanks Logitech. I am aware of the formula, but I am not sure if its relevent here.

A gives the prob of both events occuring. I would think that one EventA*EventB=prob of both events occuring.

For example, if the prob of event A is (.6) and prob of event B is (.2) is the prob of both events occuring not (.6)(.2)=(.48)?

It seems that we know the prob of both events occuring (.3) and the prob of event A occuring (.5), leaving event B to be (.6).

I've been looking over this question all day due to this issue, any comments would be appreciated!
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by logitech » Tue Jan 13, 2009 6:48 pm
720dreaming wrote:Thanks Logitech. I am aware of the formula, but I am not sure if its relevent here.

A gives the prob of both events occuring. I would think that one EventA*EventB=prob of both events occuring.

For example, if the prob of event A is (.6) and prob of event B is (.2) is the prob of both events occuring not (.6)(.2)=(.48)?

It seems that we know the prob of both events occuring (.3) and the prob of event A occuring (.5), leaving event B to be (.6).

I've been looking over this question all day due to this issue, any comments would be appreciated!
First of all,

I apologize for having you thinking on OUR careless mistake!

You are absolutely right.
These two types of probability are formulated as follows:
Probability of A and B
P(A and B) = P(A) × P(B).

In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B.

Probability of A or B
P(A or B) = P(A) + P(B).

In other words, the probability of A or B occurring is the sum of the probability of A and the probability of B.
The question stem says AND so we need use the first equation: YOUR FORMULA

And yes your 0.6 is CORRECT and A is sufficient.

So is B :)

And the answer is D
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by 720dreaming » Tue Jan 13, 2009 7:01 pm
Not a problem, your other posts/solutions MORE THAN make up for it!
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Re: Probability DS

by Ian Stewart » Wed Jan 14, 2009 12:53 am
aj5105 wrote:If the probability that event A will occur is 0.5, what is the probability that event B will occur?
(1) The probability that both A and B will occur is 0.3.
(2) The probability that at least one of the events will occur is 0.8.
This doesn't seem much like a real GMAT probability question, but since the question doesn't mention whether A and B are independent events, you have a lot less information here than you might think at first. The solution in the original post was correct; the others are not. In all of the solutions that followed the first, people have assumed that the events were *independent*. That is the only scenario when you can safely use the formula P(A AND B) = P(A)*P(B). If the events are not independent, that formula doesn't apply. It's easy to illustrate why neither statement is sufficient with a simple real world example:

Suppose you have ten cards, numbered from 1 to 10. You pick one card. Let A be "the number on the card is odd". Then the probability of A is 0.5.

If B is "the number on the card is less than 7", then there is a 0.3 chance that A and B both occur. If instead B is "the number on the card is less than 6", then there is also a 0.3 chance that A and B both occur. So 1) cannot be sufficient; the probability of B is 0.6 in the first case and 0.5 in the second case.

If B is "the number on the card is less than 7" then there is a 0.8 chance that A or B (or both) occurs. If B is "the number on the card is less than 8", there is again a 0.8 chance that A or B (or both) occurs. So 2) cannot be sufficient either.
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