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# PS: probability

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08 Oct 2003, 09:38
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There was a typo, read question as:

If P(A) = 0.83, P(B) = 0.25. If C is a set of possible values of probability that A & B both occur, Find the range of C?

Last edited by Vicky on 10 Oct 2003, 05:00, edited 1 time in total.

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Re: PS: probability [#permalink]

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08 Oct 2003, 15:04
Vicky wrote:
praet you asked for some good problems in probability, statistics...

Two events A & B are independent. If P(A) = 0.83, P(B) = 0.25.
If C is a set of possible values of probability that A & B both occur, Find the range of C?

If the events are independent .. P(A/B) = P(A) similarly ...P(B/A) = P(B)

P(A and B) = P(A ) * P (B) = 0.83 * 0.25 = 0.83/4

Is there a range for the values of C...there is just one probability...

i guess the range is 0.

thanks
praetorian

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08 Oct 2003, 22:20
seems like 0, for A and B both have to occure
The range is possible when A occures, but B does not, both does not, and so on.

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Senior Manager
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10 Oct 2003, 22:24
Folks there was a type with the question. Kindly read as:
Two events A & B are such that P(A) = 0.83, P(B) = 0.25. If C is a set of possible values of probability that A & B both occur, Find the range of C?

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10 Oct 2003, 22:49
Vicky wrote:
Folks there was a type with the question. Kindly read as:
Two events A & B are such that P(A) = 0.83, P(B) = 0.25. If C is a set of possible values of probability that A & B both occur, Find the range of C?

Ok so we dont know if the events are independent or not.

P(A/B) = P(A and B) / P(B)

C = P(A/B) * P(B)

We have two cases

1. Both events can be mutually exclusive...as in... heads on a coin and a six on a die..

P(A/B) = 1

2. Both events can be independent , in that case ...

P(A/B) = P(B) ..In case , the required probability is 0.83/4

3. Both events cannot occur together ..as in.. heads and tails in the same toss.

P(A/B) = 0

Range : Max - Min = 1 - 0 = 1 ?

thanks
praetorian

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11 Oct 2003, 00:40
praetorian123 wrote:
Vicky wrote:
Folks there was a type with the question. Kindly read as:
Two events A & B are such that P(A) = 0.83, P(B) = 0.25. If C is a set of possible values of probability that A & B both occur, Find the range of C?

Ok so we dont know if the events are independent or not.

P(A/B) = P(A and B) / P(B)

C = P(A/B) * P(B)

We have two cases

1. Both events can be mutually exclusive...as in... heads on a coin and a six on a die..

P(A/B) = 1

2. Both events can be independent , in that case ...

P(A/B) = P(B) ..In case , the required probability is 0.83/4

3. Both events cannot occur together ..as in.. heads and tails in the same toss.

P(A/B) = 0

Range : Max - Min = 1 - 0 = 1 ?

thanks
praetorian

there should be no difference between 1 and 3. Mutually exclusive events are, by definition, events that cannot happen at the same time.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Senior Manager
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11 Oct 2003, 06:48
praet try again... not the correct approach.
thanks

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12 Oct 2003, 03:21
Vicky wrote:
praet try again... not the correct approach.
thanks

Quote:
If P(A) = 0.83, P(B) = 0.25. If C is a set of possible values of probability that A & B both occur, Find the range of C?

lets try again.

P(A) + P(B) >1 , so the events are NOT mutually exclusive.

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

P(A or B) = P(A) + P(B) - C = 0.83 + 0.25 - C = 1.108 - C

Maximum value of C is when events are independent, C = P(A) * P(B) = 0.83 * 0.25

Minimum is C = 0.108 (C cant be 0 as the events are NOT mutually exclusive)

Range = (0.83 * 0.25 - 0.108) = 0.21 - 0.108 = 0.102

Range = 0.102

Vicky, could you please post more of these?
Also , you mentioned you had a tough combinations problem ...could you post a "similar " problem ?

Thanks
Praetorian

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12 Oct 2003, 21:47
very good.. u are on target..but missed it slightly.
First, a silly mistake in calculation
P(A or B or both) = 1.08 - P(A & B)
It is correct that Minimum of P(A&B) = .08

praet why do u say that max of P(A & B) occurs when events are independent? Independent events just mean that probaility of one does not affect that of other i.e. outcome of second event is idependent of outcome of first and vice-versa.

Okay, Now how do u find the maximum?
Can u use set theory. Just draw two vien diagram of A & B, knowing that they intersect. Can u find out for what set A & B intersection will be Max.
thanks.

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12 Oct 2003, 22:29
Vicky wrote:
very good.. u are on target..but missed it slightly.
First, a silly mistake in calculation
P(A or B or both) = 1.08 - P(A & B)
It is correct that Minimum of P(A&B) = .08

praet why do u say that max of P(A & B) occurs when events are independent? Independent events just mean that probaility of one does not affect that of other i.e. outcome of second event is idependent of outcome of first and vice-versa.

Okay, Now how do u find the maximum?
Can u use set theory. Just draw two vien diagram of A & B, knowing that they intersect. Can u find out for what set A & B intersection will be Max.
thanks.

Thanks...minimum is 0.08.
yeah, i was totally wrong.

Maximum will occur when B is a subset of A.
In that case , C =0.25
Maximum C = 0.25

Range..0.25 - 0.08 = 0.17

Hope i am right this time.

Do you have more questions like these ?

Thanks
Praetorian

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13 Oct 2003, 04:48
There u go... good. a simple but conceptual question.
i invented this question after reading prob. notes of OG.
will see if i can come up with more.
thanks

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13 Oct 2003, 04:48
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# PS: probability

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