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What is the probability that Lee will make exactly 5 errors on a certa

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New post 04 Aug 2018, 09:42
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What is the probability that Lee will make exactly 5 errors on a certain typing test?

(1) The probability that Lee will make 5 or more errors on the test is 0.27.
(2) The probability that Lee will make 5 or fewer errors on the test is 0.85.



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(DS06810)

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What is the probability that Lee will make exactly 5 errors on a certa  [#permalink]

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New post Updated on: 04 Aug 2018, 13:29
2
Bunuel wrote:
What is the probability that Lee will make exactly 5 errors on a certain typing test?


(1) The probability that Lee will make 5 or more errors on the test is 0.27.
Exactly 5 errors + More than 5 errors = 0.27
No information about fewer than 5 errors
Insufficient

(2) The probability that Lee will make 5 or fewer errors on the test is 0.85.
Fewer than 5 errors + Exactly 5 errors = 0.85
No information about More than 5 errors
Insufficient

Combining both :
(=> 5 errors) + (=< 5 errors)
= 0.27 + 0.85
= 1.12
Now, notice how a probability can not be more than 1, so the extra value obtained i.e 0.12 must be that of the exact 5 errors.
Therefore, 1.12 - 1 = 0.12
Sufficient

Hence, C.
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Originally posted by sudarshan22 on 04 Aug 2018, 11:10.
Last edited by sudarshan22 on 04 Aug 2018, 13:29, edited 1 time in total.
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What is the probability that Lee will make exactly 5 errors on a certa  [#permalink]

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New post 04 Aug 2018, 11:23
2
sudarshan22 wrote:
Bunuel wrote:
What is the probability that Lee will make exactly 5 errors on a certain typing test?


(1) The probability that Lee will make 5 or more errors on the test is 0.27.
Exactly 5 errors + More than 5 errors = 0.27
No information about fewer than 5 errors
Insufficient

(2) The probability that Lee will make 5 or fewer errors on the test is 0.85.
Fewer than 5 errors + Exactly 5 errors = 0.85
No information about More than 5 errors
Insufficient

Combining both :
(=> 5 errors) + (=< 5 errors)
= 0.27 + 0.85
= 1.12
Now, notice how a probability can not be more than 1, so the extra value obtained i.e 0.12 must be that of the exact 5 errors.
And since the probability of exact 5 errors is added twice while combining both statements(once from each statement), half of the extra value obtained must be the probability of exactly 5 errors on a certain typing test

Therefore, 1.12 - 1 = 0.12
Half of 0.12 = 0.06
Sufficient


Hence, C.

sudarshan22
Just small correction

P(5)+P(6)+............+P(n) = 0.27 ......Statement 1
P(0)+P(1)+ P(2)+........+P(5) = 0.85...Statement 2
Adding statement 1 and statement 2
[P(0)+P(1)................+P(5)+...........+P(n)] + P(5) = 0.27+0.85
1 +P(5) = 1.12
P(5)= 1.12-1 =0.12
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What is the probability that Lee will make exactly 5 errors on a certa  [#permalink]

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New post 04 Aug 2018, 11:43
Bismarck wrote:
Just small correction

P(5)+P(6)+............+P(n) = 0.27 ......Statement 1
P(0)+P(1)+ P(2)+........+P(5) = 0.85...Statement 2
Adding statement 1 and statement 2
[P(0)+P(1)................+P(5)+...........+P(n)] + P(5) = 0.27+0.85
1 +P(5) = 1.12
P(5)= 1.12-1 =0.12


Bismarck
Ohh yes, that makes much sense. I was just plugging and chugging and what not to solve the question.
Glad it was just a DS question(exact value is not required), and not a PS question.

Thanks much for rectifying the blunder . :thumbup:
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What is the probability that Lee will make exactly 5 errors on a certa  [#permalink]

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New post 04 Aug 2018, 13:23
Bunuel wrote:
What is the probability that Lee will make exactly 5 errors on a certain typing test?

(1) The probability that Lee will make 5 or more errors on the test is 0.27.
(2) The probability that Lee will make 5 or fewer errors on the test is 0.85.

NEW question from GMAT® Quantitative Review 2019


(DS06810)


Let P(A)=Probability that Lee will make 5 or more errors on the test.
P(B)=Probability that Lee will make 5 or fewer errors on the test.
We have, P(A or B)=P(A)+P(B)-P(A and B)------------(1)

Question stem:- Probability that Lee will make exactly 5 errors on a certain typing test=P(A and B)=?
From (1), we have P(A and B)=P(A)+P(B)-P(A or B)---------------(2)

St1:- The probability that Lee will make 5 or more errors on the test is 0.27.
Or, P(A)=0.27.
We can't determine P(A and B).
Insufficient.

St2:- The probability that Lee will make 5 or more errors on the test is 0.27.
Or, P(B)=0.85.
We can't determine P(A and B).
Insufficient.

Combining, we have P(A or B)=1 [Lee makes at least 5 errors or at most 5 errors]
P(A and B)=P(A)+P(B)-P(A or B)
Or, P(A and B)=0.27+0.85-1=1.12-1=0.12
Sufficient.

Ans. (C)
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Re: What is the probability that Lee will make exactly 5 errors on a certa  [#permalink]

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New post 05 Aug 2018, 05:39
Bunuel wrote:
What is the probability that Lee will make exactly 5 errors on a certain typing test?

(1) The probability that Lee will make 5 or more errors on the test is 0.27.
(2) The probability that Lee will make 5 or fewer errors on the test is 0.85.

NEW question from GMAT® Quantitative Review 2019


(DS06810)


A good question.

a. prob of 5 errors + prob of more than 5 errors = .27 ---> NS

b. prob of 5 errors + prob of less than 5 errrors= .85 ---> NS

both of them individually give no idea of exactly 5 errors..

we get 2 equations on combining both so C
Re: What is the probability that Lee will make exactly 5 errors on a certa &nbs [#permalink] 05 Aug 2018, 05:39
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