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# Probability

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13 Jul 2007, 08:04
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Xavier, Yvonne and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2, and 5/8 respectively, what is the probability that Xavier and Yvonne but not Zelda, will solve the problem?

a. 11/8
b. 7/8
c. 9/64
d. 5/64
e. 3/64

I read the tutorial this forum offers, the NOT tool says:
p(not A) + p(A) = 1. I tried it but I could not come up with the answer. Thank you for helping.
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13 Jul 2007, 08:39
The probability of Xavier solve the problem is 1/4
the probability of Yvonne solve the problem is 1/2
the probability of Zelda not solve the problem is 1-5/8, or 3/8
So the probability of Xavier and Yvonne but not Zelda solve the problem will be 1/4*1/2*3/8, or 3/64. Answer E
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13 Jul 2007, 08:48
What is the probability of Xavier or Yvonne but not Zelda solving the problem?
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13 Jul 2007, 09:09
Using the OR tool: p(A or B) = p(A) + p(B) - p(A and B).

1/4+1/2- (1/4*1/2)=1/2.

Then I subtracted Zelda's probability of NOT answering correctly (3/8).

1/2-3/8=3/16

Correct me if I'm wrong please. I started trying these probability problems only recently.
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13 Jul 2007, 11:12
joselord wrote:
Using the OR tool: p(A or B) = p(A) + p(B) - p(A and B).

1/4+1/2- (1/4*1/2)=1/2.

Then I subtracted Zelda's probability of NOT answering correctly (3/8).

1/2-3/8=3/16

Correct me if I'm wrong please. I started trying these probability problems only recently.

You have to multiply 1/2 and 3/8 Not subtract
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13 Jul 2007, 11:19
dahcrap wrote:
joselord wrote:
Using the OR tool: p(A or B) = p(A) + p(B) - p(A and B).

1/4+1/2- (1/4*1/2)=1/2.

Then I subtracted Zelda's probability of NOT answering correctly (3/8).

1/2-3/8=3/16

Correct me if I'm wrong please. I started trying these probability problems only recently.

You have to multiply 1/2 and 3/8 Not subtract

I think he is multiplying
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13 Jul 2007, 12:29
boubi wrote:
What is the probability of Xavier or Yvonne but not Zelda solving the problem?

1. X & Y but not Z = 1/4*1/2 *3/8 = 3/64

2. X and not Y and not Z = 1/4 * 1/2 * 3/8 = 3/64

3. not X and Y and not Z = 3/4*1/2*3/8 = 9/64

Adding 1, 2 and 3 = 13/64
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13 Jul 2007, 14:45
What is the probability that either Xavier or Yvonne but not Zelda solving the problem?
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13 Jul 2007, 15:34
asaf wrote:
boubi wrote:
What is the probability of Xavier or Yvonne but not Zelda solving the problem?

1. X & Y but not Z = 1/4*1/2 *3/8 = 3/64

2. X and not Y and not Z = 1/4 * 1/2 * 3/8 = 3/64

3. not X and Y and not Z = 3/4*1/2*3/8 = 9/64

Adding 1, 2 and 3 = 13/64

Is this right? I calculated it as:

Prob of X or Y= (1/4)+(1/2)-(1/4)(1/2)=(6/8)-(1/8)=5/8. Multiply by the probability that Zelda doesn't solve: (5/8)*(3/8)=15/64.

I think you just added 1, 2, and 3 incorrectly.
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13 Jul 2007, 16:04
briks123 wrote:
asaf wrote:
boubi wrote:
What is the probability of Xavier or Yvonne but not Zelda solving the problem?

1. X & Y but not Z = 1/4*1/2 *3/8 = 3/64

2. X and not Y and not Z = 1/4 * 1/2 * 3/8 = 3/64

3. not X and Y and not Z = 3/4*1/2*3/8 = 9/64

Adding 1, 2 and 3 = 13/64

Is this right? I calculated it as:

Prob of X or Y= (1/4)+(1/2)-(1/4)(1/2)=(6/8)-(1/8)=5/8. Multiply by the probability that Zelda doesn't solve: (5/8)*(3/8)=15/64.

I think you just added 1, 2, and 3 incorrectly.

I dont think you can subtract (1/4) * (1/2). The assumption here is independent events. I may be wrong as I am having a horrible day today
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13 Jul 2007, 21:04
dahcrap wrote:
briks123 wrote:
asaf wrote:
boubi wrote:
What is the probability of Xavier or Yvonne but not Zelda solving the problem?

1. X & Y but not Z = 1/4*1/2 *3/8 = 3/64

2. X and not Y and not Z = 1/4 * 1/2 * 3/8 = 3/64

3. not X and Y and not Z = 3/4*1/2*3/8 = 9/64

Adding 1, 2 and 3 = 13/64

Is this right? I calculated it as:

Prob of X or Y= (1/4)+(1/2)-(1/4)(1/2)=(6/8)-(1/8)=5/8. Multiply by the probability that Zelda doesn't solve: (5/8)*(3/8)=15/64.

I think you just added 1, 2, and 3 incorrectly.

I dont think you can subtract (1/4) * (1/2). The assumption here is independent events. I may be wrong as I am having a horrible day today

dahcrap, dont beat up on yourself so much - you hav solved so many tough problems - i surely think u'll score in the 99th+ percentile.

anywya, my understanding is just the opposite
firstly, that you subtract Px + Py by Pxy bcoz they are independent (whereas, if they are mutually exclusive, there is nothing to subtract)
in other words, any two or all of them can solve the problem.

solution of asaf and briks seems correct to me (but of course, i am not an expert).
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13 Jul 2007, 21:23
oops wrote:
dahcrap wrote:
briks123 wrote:
asaf wrote:
boubi wrote:
What is the probability of Xavier or Yvonne but not Zelda solving the problem?

1. X & Y but not Z = 1/4*1/2 *3/8 = 3/64

2. X and not Y and not Z = 1/4 * 1/2 * 3/8 = 3/64

3. not X and Y and not Z = 3/4*1/2*3/8 = 9/64

Adding 1, 2 and 3 = 13/64

Is this right? I calculated it as:

Prob of X or Y= (1/4)+(1/2)-(1/4)(1/2)=(6/8)-(1/8)=5/8. Multiply by the probability that Zelda doesn't solve: (5/8)*(3/8)=15/64.

I think you just added 1, 2, and 3 incorrectly.

I dont think you can subtract (1/4) * (1/2). The assumption here is independent events. I may be wrong as I am having a horrible day today

dahcrap, dont beat up on yourself so much - you hav solved so many tough problems - i surely think u'll score in the 99th+ percentile.

anywya, my understanding is just the opposite
firstly, that you subtract Px + Py by Pxy bcoz they are independent (whereas, if they are mutually exclusive, there is nothing to subtract)
in other words, any two or all of them can solve the problem.

solution of asaf and briks seems correct to me (but of course, i am not an expert).

on the other hand, in the case that they are mutually exclusive, the probability for X and Y are added, but P for Z is subtracted instead of multiplying by NOT Z.

for example, here is a question from gmatclub Challenge 7:
Probability Mike can win a championship is 1/4, Rob 1/3 and Ben 1/6. What is the Probability that Mike or Rob win but not Ben?

the answer to this from my notes: 1/4 + 1/3 - 1/6 = 5/12
my understanding is this is bcoz only one of them can win the championship, unlike the previous problem.

probability gurus, plz comment.
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13 Jul 2007, 23:12
oops wrote:
for example, here is a question from gmatclub Challenge 7:

Probability Mike can win a championship is 1/4, Rob 1/3 and Ben 1/6. What is the Probability that Mike or Rob win but not Ben?

i would solve this problem this way:

= (1/4 + 1/3) (1 - 1/6)
=(7/12)(5/6)
= 35/72
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13 Jul 2007, 23:13
joselord wrote:
Xavier, Yvonne and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2, and 5/8 respectively, what is the probability that Xavier and Yvonne but not Zelda, will solve the problem?

a. 11/8
b. 7/8
c. 9/64
d. 5/64
e. 3/64

= 1/4 x 1/2 x (1 - 5/8)
= 3/64
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14 Jul 2007, 05:17
These are independent events.
So the probability = 1/4 * 1/2 * 3/8 = 3/64

Release the OA please.
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14 Jul 2007, 05:39
Fistail wrote:
oops wrote:
for example, here is a question from gmatclub Challenge 7:

Probability Mike can win a championship is 1/4, Rob 1/3 and Ben 1/6. What is the Probability that Mike or Rob win but not Ben?

i would solve this problem this way:

= (1/4 + 1/3) (1 - 1/6)
=(7/12)(5/6)
= 35/72

you are right. thanks for the input.
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14 Jul 2007, 15:21
briks123 wrote:
Fistail wrote:
oops wrote:
for example, here is a question from gmatclub Challenge 7:

Probability Mike can win a championship is 1/4, Rob 1/3 and Ben 1/6. What is the Probability that Mike or Rob win but not Ben?

i would solve this problem this way:

= (1/4 + 1/3) (1 - 1/6)
=(7/12)(5/6)
= 35/72

you are right. thanks for the input.

as mentioned in the post, the OA given is 5/12 - this problem/solution is in GMATClub Challenge No. 7
[#permalink] 14 Jul 2007, 15:21
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# Probability

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