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# Xavier, Yvonne, and Zelda each try independently to solve a

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Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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20 Jun 2008, 00:58
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Oct 2012, 03:56, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.
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20 Jun 2008, 01:10
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OG C# 231
Xavier, Yvonne, Zelda each try independently to solve a problem. If their individual probabilities
for success are 1/4, 1/2, and 5/8, repectively, what is the probability that xavier and yvonne, but
not zelda, will solve the problem? Please provide solution with explanation. THanks
a. 11/8
b. 7/8
c. 9/64
d. 5/64
e. 3/64

P(Xavier will solve)=1/4
P(Yvonne will solve)=1/2
P(Zelda will NOT solve) = 1- 5/8 = 3/8.

Now, we need to multiply all this Ps to find an answer:
p= (1/4)*(1/2)*(3/8) = 3/64.

Ans. E.
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Re: OG C# 231 Xavier, Yvonne, Zelda each try independently to [#permalink]

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12 Oct 2012, 09:36
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Yes - since they are independent events, you can multiply the probabilities together.

Be sure to remember to take the opposite probability for that last one.

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02 May 2013, 15:46
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greenoak wrote:
Quote:
OG C# 231

Now, we need to multiply all this Ps to find an answer:
p= (1/4)*(1/2)*(3/8) = 3/64.

I always get confused, Why don't we multiply it by 3? because these independent event can occur is any order. that means it can also be (3/8)*(1/4)*(1/2) ?
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16 May 2013, 14:53
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I'm not great with probability so double check whatever I say, but we're not looking for the order in which X and Y get it right and Z get's it wrong. We're just looking for the probability of X and Y getting it right while Z get's it wrong. It doesn't matter in what order they get it right or wrong, all that matters is how they perform. Think of it like flipping a coin a few times and recording the probability of getting heads. It doesn't matter if you get two heads in a row then one tails, or one tails then two heads in a row. The probability is the same either way. This is in contrast to say, the chances of gettign a certain color gum ball in a certain order.

Hope that helps!

nikhil007 wrote:
greenoak wrote:
Quote:
OG C# 231

Now, we need to multiply all this Ps to find an answer:
p= (1/4)*(1/2)*(3/8) = 3/64.

I always get confused, Why don't we multiply it by 3? because these independent event can occur is any order. that means it can also be (3/8)*(1/4)*(1/2) ?
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Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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22 Oct 2014, 09:15
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Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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21 Dec 2015, 02:29
Hello from the GMAT Club BumpBot!

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Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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23 Dec 2015, 12:39
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Hi All,

In this type of probability question, we're asked for a specific outcome. To solve this problem, we'll have to deal with each piece individually, then multiply the outcomes together.

Xavier solves the problem
Yvonne solves the problem
Zelda does NOT solve the problem.

Xavier's probability to solve = 1/4
Yvonne's probability to solve = 1/2
Zelda's probability to NOT solve = 1 - 5/8 = 3/8

The final answer is (1/4)(1/2)(3/8) = 3/64

[Reveal] Spoiler:
E

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Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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15 Jul 2016, 04:37
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

We are first given the individual probabilities that Xavier, Yvonne, and Zelda WILL solve the problem. We list these below:

P(Xavier will solve) = ¼

P(Yvonne will solve) = ½

P(Zelda will solve) = 5/8

However, we see the question asks for the probability that Xavier and Yvonne, but not Zelda, will solve the problem.
Thus we must determine the probability that Zelda WILL NOT solve the problem. "Solving" and "not solving" are complementary events. When two events are complementary, knowing the probability that one event will occur allows us to calculate the probability that the complement will occur. That is, P(A) + P(Not A) = 1. In the case of Zelda, the probability that she WILL NOT solve the problem, is the complement of the probability that she WILL solve the problem.

P(Zelda will solve) + P(Zelda will not solve) = 1

5/8 + P(Zelda will not solve) = 1

P(Zelda will not solve) = 1 – 5/8 = 3/8

Now we can determine the probability that Xavier and Yvonne, but not Zelda, will solve the problem. Since we need to determine three probabilities that all must take place, we must multiply the probabilities together. Thus we have:

P(Xavier will solve) x P(Yvonne will solve) x P(Zelda will not solve)

¼ x ½ x 3/8

1/8 x 3/8 = 3/64

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Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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31 Jan 2017, 17:45
P(X and Y) Solving Problem = 1/4 * 1/2 = 1/8
P(Z) NOT Solving Problem = 1 - 5/8 = 3/8
P(X and Y not Z) = 1/8*3/8
P = 3/64
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Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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18 May 2017, 09:43
P(X) = 1/4
P(Y) = 1/2
P(z) = 5/8

P(X and Y and NOT Z)= ?

Probability of something happening = 1 – P (of something not happening)

P (not Z) = 1 – 5/8 = 3/8
P(X and Y and NOT Z)= 1/4*1/2*3/8 = 3/64

Re: Xavier, Yvonne, and Zelda each try independently to solve a   [#permalink] 18 May 2017, 09:43
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