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A student is to take her final exams in two subjects. The probability

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A student is to take her final exams in two subjects. The probability  [#permalink]

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Tough and Tricky questions: Probability.



A student is to take her final exams in two subjects. The probability that she will pass the first subject is 3/4 and the probability that she will pass the second subject is 2/3. What is the probability that she will pass one exam or the other exam?

A) 5/12
B) 1/2
C) 7/12
D) 5/7
E) 11/12

Kudos for a correct solution.

Source: Chili Hot GMAT

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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 05 Dec 2014, 07:30
Let the two subjects be X (probability of passing 3/4) and Y (probability of passing 2/3)
The probability of passing in X only is 3/4 times 1/3 (failing in Y) = 1/4
The probability of passing in Y only is 2/3 times 1/4 (failing in X) = 1/6
Therefore the total probability = 1/4 + 1/6 = 5/12

Hence A
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 05 Dec 2014, 09:27
We need to compute two different probabilities here--the probability that she passes the first exam and fails the second exam and the probability that she passes the second exam and fails the first exam. The probability of failing an exam is 1 - the probability of passing the exam. Then sum the probabilities for your answer:

P = 3/4*1/3 + 1/4*2/3 = 1/4 + 1/6 = 5/12

Choice A
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 05 Dec 2014, 17:40
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Do we really know that P(x) means only passing in x? If we visualize the venn diagram, P(x) means probability of passing in x along with yand without y!
My answer would be: 1/3+1/4-1/3*1/4=11/12
This is P(XUY).
Answer E.
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 05 Dec 2014, 17:43
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1
Do we really know that P(x) means only passing in x? If we visualize the venn diagram, P(x) means probability of passing in x along with yand without y!
My answer would be: 1/3+1/4-1/3*1/4=11/12
This is P(XUY).
Answer E.
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 05 Dec 2014, 18:42
Hi farzana87,

Probabilities are given for P(x) as a single event; and not in conjunction to any other event. So when you try to put it in a venn diagram you are actually trying to find out the parts xNOTy + yNOTx.

nb: Venn diagrams may not be the best way to help you with such problems! Bujhle?
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 05 Dec 2014, 22:51
:( Yes I found out that P(XUY) means passing in one or two subjects. omegacube
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 06 Dec 2014, 00:57
A student is to take her final exams in two subjects. The probability that she will pass the first subject is 3/4 and the probability that she will pass the second subject is 2/3. What is the probability that she will pass one exam or the other exam?

A) 5/12
B) 1/2
C) 7/12
D) 5/7
E) 11/12


Probability that the student will pass the first subject is 3/4 -> Probability that she will fail in the first subject is 1/4
Probability that she will pass the second subject is 2/3 -> Probability that she will fail in the second subject is 1/3

Probability that she will pass one exam OR the other exam
=Probability that she will pass the first subject AND fail in the second subject + Probability that she will pass the second subject AND fail in the first subject

= \((\frac{3}{4})(\frac{1}{3}) + (\frac{1}{4})(\frac{2}{3}) =\frac{1}{4}+ \frac{1}{6} = \frac{5}{12}\)

Answer: A
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 06 Dec 2014, 11:04
answer should be A.

---- pass fail
1: 3/4 1/4
2: 2/3 1/3

3/4*1/3 + 2/3*1/4

=5/12
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 06 Dec 2014, 12:21
This question can also be solved by using venn diagram.

Image

As show in the figure above. the required probability is the highlighted area in red. let circle A represents probability of passing in first exam =3/4, and circle B represents probability of passing in second exam = 2/3 and the probability of passing in both the exams = (2/3)(3/4) = 1/2

thus highlighted portion on the left hand side can be calculated as 3/4-1/2 = 1/4
highlighted portion on the right hand side can be calculated as 2/3-1/2 = 1/6

thus required probability is 1/4+1/6 = 5/12
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 06 Dec 2014, 21:18
A student is to take her final exams in two subjects. The probability that she will pass the first subject is 3/4 and the probability that she will pass the second subject is 2/3. What is the probability that she will pass one exam or the other exam?

Alternate Solution (Table Method)
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 18 Mar 2015, 11:01
(3/4 +2/3)-3/4*2/3 = 11/12
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 18 Mar 2015, 20:56
1
In probability for multiple event experiments, the probability for A or B is the following:

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

So we substitute the data from the question:

3/4 + 2/3 - (3/4 * 2/3) = 17/12 - 6/12 = 11/12

Answer is E
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 19 Mar 2015, 06:07
Bunuel wrote:
Bunuel wrote:

Tough and Tricky questions: Probability.



A student is to take her final exams in two subjects. The probability that she will pass the first subject is 3/4 and the probability that she will pass the second subject is 2/3. What is the probability that she will pass one exam or the other exam?

A) 5/12
B) 1/2
C) 7/12
D) 5/7
E) 11/12

Kudos for a correct solution.

Source: Chili Hot GMAT


The correct answer is E.


Hi Bunuel

can you pl let me know where am I going wrong? i got 7/12. please refer to the image...

i got 1/3 + 1/4 = 7/12 :(

is this because i am considering that failing in both is not an option?
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 19 Mar 2015, 06:20
this is a typical question that "begs" you to identify the probability of not passing either of the exams.
In this particular case, the probability of not passing the first exam is 1/4. The probability of not passing the second exam is 1/3.
The probability of not passing both of the exams is thus equal to (1/4) * (1/3) = 1/12
We can then deduct 1/12 from one, and we will get 11/12. This is the probability of passing at least one exam :)
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 19 Mar 2015, 18:49
I derived the answer as follows:

P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 3/4 + 2/3 - 1/2
P(A or B) = 11/12

Any suggestions on the correctness of approach? Please guide.
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 19 Mar 2015, 19:20
HI All,

The "intent" of this question is not perfectly clear from the wording.

IF the intent is to ask....what is the probability of passing JUST ONE test, then the answer is 5/12.

IF the intent is to ask....what is the probability of passing AT LEAST ONE test, then the answer is 11/12.

The various approaches for those solutions have already been presented by other posters, so I won't rehash any of that math here. The Official GMAT writes (and tests out) it's questions so that the meaning is clear and there is no 'interpretational bias.', so Test Takers won't left to question what the prompt actually asks for.

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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 21 May 2016, 12:35
EMPOWERgmatRichC wrote:
HI All,

The "intent" of this question is not perfectly clear from the wording.

IF the intent is to ask....what is the probability of passing JUST ONE test, then the answer is 5/12.

IF the intent is to ask....what is the probability of passing AT LEAST ONE test, then the answer is 11/12.

The various approaches for those solutions have already been presented by other posters, so I won't rehash any of that math here. The Official GMAT writes (and tests out) it's questions so that the meaning is clear and there is no 'interpretational bias.', so Test Takers won't left to question what the prompt actually asks for.

GMAT assassins aren't born, they're made,
Rich


I too had the same doubt when solving the question. Doubt is that whether it is asking that the student passes EXACTLY one exam then the answer would be 5/12 or the student needs to pass at least one of the exams in which case the answer is 11/12
I think the question stem is asking for first approach but wonder why the correct answer is given for latter approach
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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New post 22 May 2016, 11:42
probability of passing neither=(1/3)(1/4)=1/12
probability of passing either=1-1/12=11/12
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Re: A student is to take her final exams in two subjects. The probability  [#permalink]

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