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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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New post Updated on: 13 Oct 2012, 03:56
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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

Originally posted by redbeanaddict on 20 Jun 2008, 00:58.
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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Re: Xavier, Yvonne, and Zelda each try independently to solve a  [#permalink]

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New post 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.

We're asked for 3 things:

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

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Re: OG C 231  [#permalink]

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New post 20 Jun 2008, 01:10
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Quote:
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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New post 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.

Please see: http://www.gmatpill.com/gmat-practice-t ... stion/2387

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Re: OG C 231  [#permalink]

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New post 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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Re: OG C 231  [#permalink]

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New post 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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New post 15 Jul 2016, 04:37
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redbeanaddict 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


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

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

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New post 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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New post 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

The answer is E.
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Re: Xavier, Yvonne, and Zelda each try independently to solve a  [#permalink]

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New post 22 Jul 2017, 08:33
EMPOWERgmatRichC wrote:
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.

We're asked for 3 things:

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

Final Answer:

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Hi Rich,

Could you please tell me why dont we multiply this by 3! ?
(X)(Y)(not Z)
(not Z)(X)(Y)....so on till 6 times.

Please clarify.

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

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New post 22 Jul 2017, 12:26
Hi Sash143,

The question does not state anything about what the "order" of the outcomes must be - just the probability of each of the individual outcomes (and the specific 'overall' outcome that we're looking to calculate). Thus, it's not a permutation question (meaning that it does NOT matter who attempts to solve the problem first, second or third) - and we don't have to do anything besides multiply the individual probabilities together to get the correct answer.

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

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New post 24 Jul 2017, 12:14
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redbeanaddict 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


P(Z solves the problem) = 1 - P(Z doesn't solve the problem)
So, 5/8 = 1 - P(Z doesn't solve the problem)
So, P(Z doesn't solve the problem) = 3/8

The question asks us to find P(Xavier and Yvonne solve problem, but Zelda does not solve problem)
So, we want: P(X solves problem AND Y solves problem AND Z does not solve)
= P(X solves problem) x P(Y solves problem) x P(Z does not solve)
= 1/4 x 1/2 x 3/8
= 3/64

Answer:

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

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New post 05 Nov 2017, 01:09
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if x can solve with probability of 25 % and Y can at 50 % then combined together how come it is lesser than both????????? ie here 1/8.. i cannot feel this physically . could somebody explain this??
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Xavier, Yvonne, and Zelda each try independently to solve a  [#permalink]

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New post 08 Nov 2017, 10:42
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Cheryn wrote:
if x can solve with probability of 25 % and Y can at 50 % then combined together how come it is lesser than both????????? ie here 1/8.. i cannot feel this physically . could somebody explain this??

Cheryn, maybe what follows will help, because intuition is important: Think about how strictly or restrictively "success" is defined. (Success = desired outcome = win = passing the test.)

The probability of "success" here is restrictive because BOTH have to pass. If only X OR Y had to pass, success would be easier.

To win, "BOTH this AND that must happen." That is a stricter definition of success than "to win, EITHER this OR that must happen."

In the first case, two people must succeed. In the second case (OR), only one person OR the other person must succeed. It's harder to get two wins than it is to get one win.

Look at the difference: if the question were "What is the probability of X or Y passing the test?" The answer:
\(\frac{1}{2} + \frac{1}{4}=\frac{3}{4}\)
One OR the other must pass? Easier, less restrictive than "both must pass."

Different scenario, but it works exactly the same way.

A coin toss. What is the probability that one coin, flipped twice, will land on tails both times? Success = tails on the first flip AND on the second flip

P (tails) on the first flip is \(\frac{1}{2}\)
P (tails) on second flip = \(\frac{1}{2}\)

Events are independent.
Multiply:\(\frac{1}{2}* \frac{1}{2}=\frac{1}{4}\)

The probability of having both flips come up tails, \(\frac{1}{4}\), is lower (smaller) than the probability of having just one flip come up tails (\(\frac{1}{2}\)).

Again, that is because success is defined more restrictively. It is harder to get two tails on two flips than it is to get one tail on two flips; you have to "beat the odds" twice, not once. Lower probability.

Most probabilities are fractions between 0 and 1. When those fractions are multiplied, they get smaller. That fits.

It can be a little counterintuitive if you focus on AND. "And" might seem as if it should produce a better chance of success than "or." Maybe focus instead on: the definition of success, and how success is achieved.

Almost always, (BOTH must win) will be harder (lower probability) than (ONE OR THE OTHER must win).

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

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New post 09 Nov 2017, 02:05
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 to get answer
p= (1/4)*(1/2)*(3/8) = 3/64.
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Re: Xavier, Yvonne, and Zelda each try independently to solve a  [#permalink]

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New post 10 Nov 2017, 10:40
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thanks genxter123 it do helped me to understand, especially initial part. thanks again
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Re: Xavier, Yvonne, and Zelda each try independently to solve a  [#permalink]

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New post 07 Jan 2018, 06:58
redbeanaddict 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


Greetings :-)

can anyone explain whats wrong with the following solution:

multiplied 1/4 * 1/2 * 5/8 = 5/64

than i thought we need to subtract 5/8 from 5/64

isnt it locically correct ? :?

thank you ! :)
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Xavier, Yvonne, and Zelda each try independently to solve a  [#permalink]

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New post 07 Jan 2018, 07:09
dave13 wrote:
redbeanaddict 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


Greetings :-)

can anyone explain whats wrong with the following solution:

multiplied 1/4 * 1/2 * 5/8 = 5/64

than i thought we need to subtract 5/8 from 5/64

isnt it locically correct ? :?

thank you ! :)


Hi dave13!

Were you going for an approach of calculating 1 - the probability?
In that case you would have needed to calculate (Xavier and Yvonne succeed) - (Xavier and Yvonne and Zelda succeed).
Which is to say 1/4*1/2 - 1/4 * 1/2 * 5/8 = 1/8 - 5/64 = 3/64, answer (E).

What you wrote above was (Xavier and Yvonne and Zelda succeed) - (Zelda succeeds).
This doesn't make sense because it is negative, but even if it were reversed then it would have given you a different result:
(Zelda succeeds) - (Xavier and Yvonne and Zelda succeed) is the probability that (Zelda succeeds and (one or more of Xavier and Yvonne fail))

Pay attention to what is your "1" and what is your "probability" when you calculate "1 - probability"!

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

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New post 07 Jan 2018, 07:51
Since their individual probabilities for success are
p(x) = 1/4,
p(y)= 1/2 and
p(z)= 5/8, respectively, => failure probability p'(z) = 1- p(z)
the probability that Xavier and Yvonne, but not Zelda, will solve the problem = p(x)*p(y)*p'(z)
=(1/4)*(1/2)*(1-5/8)
=3/64
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Re: Xavier, Yvonne, and Zelda each try independently to solve a  [#permalink]

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New post 18 Jan 2018, 01:33
P(Xavier )=1/4
P(Yvonne)=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: Xavier, Yvonne, and Zelda each try independently to solve a   [#permalink] 18 Jan 2018, 01:33

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