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505-555 Level|   Probability|                     
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Bunuel
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What is the probability that Lee will make exactly 5 errors on 04 Aug 2018, 09:42
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C
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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.


ID: 700311
(DS06810)­
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C
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What is the probability that Lee will make exactly 5 errors on 07 Oct 2018, 07:48
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Bunuel
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)
Show more


We know that

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors) ....... (I)

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

P (Exactly 5 errors) + P (More than 5 errors) = 0.27
Putting this in (I) we can get the value of P (Fewer than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

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

P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.85
Putting this in (I) we can get the value of P (More than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

Using both,
P (Exactly 5 errors) + P (More than 5 errors) + P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.27 + 0.85
P (Exactly 5 errors) + 1 = 1.12
P (Exactly 5 errors) = 0.12
Sufficient

Answer (C)
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What is the probability that Lee will make exactly 5 errors on Updated on: 04 Aug 2018, 13:29
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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Bunuel
What is the probability that Lee will make exactly 5 errors on a certain typing test?
Show more

(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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What is the probability that Lee will make exactly 5 errors on 04 Aug 2018, 11:23
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sudarshan22
Bunuel
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.
Show more
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 04 Aug 2018, 11:43
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Bismarck
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
Show more

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 04 Aug 2018, 13:23
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Bunuel
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)
Show more

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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What is the probability that Lee will make exactly 5 errors on 05 Aug 2018, 05:39
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Bunuel
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)
Show more

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
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What is the probability that Lee will make exactly 5 errors on 05 Nov 2019, 10:19
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VeritasKarishma
Bunuel
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)


We know that

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors) ....... (I)

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

P (Exactly 5 errors) + P (More than 5 errors) = 0.27
Putting this in (I) we can get the value of P (Fewer than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

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

P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.85
Putting this in (I) we can get the value of P (More than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

Using both,
P (Exactly 5 errors) + P (More than 5 errors) + P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.27 + 0.85
P (Exactly 5 errors) + 1 = 1.12
P (Exactly 5 errors) = 0.12
Sufficient

Answer (C)
Show more


Hi karishma ,

Could you please explain why P(Exact 5) is not 0.12/2 = 0.06 as it is added twice VeritasKarishma

Posted from my mobile device
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What is the probability that Lee will make exactly 5 errors on 05 Nov 2019, 18:28
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sudarshan22
Bunuel
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.
Show more

Why the P(exact 5) is not 0.12/2 = 0.06 as it is added twice

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What is the probability that Lee will make exactly 5 errors on 06 Nov 2019, 02:36
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LeenaSai
VeritasKarishma
Bunuel
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)


We know that

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors) ....... (I)

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

P (Exactly 5 errors) + P (More than 5 errors) = 0.27
Putting this in (I) we can get the value of P (Fewer than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

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

P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.85
Putting this in (I) we can get the value of P (More than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

Using both,
P (Exactly 5 errors) + P (More than 5 errors) + P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.27 + 0.85
P (Exactly 5 errors) + 1 = 1.12
P (Exactly 5 errors) = 0.12
Sufficient

Answer (C)


Hi karishma ,

Could you please explain why P(Exact 5) is not 0.12/2 = 0.06 as it is added twice VeritasKarishma

Posted from my mobile device
Show more

Note that one P(Exactly 5 errors) is included in 1. So it is already accounted for.

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors) ....... (I)

P (Exactly 5 errors) + P (More than 5 errors) + P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.27 + 0.85
P (Exactly 5 errors) + 1 = 1.12

Hence you do not divide by 2.
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What is the probability that Lee will make exactly 5 errors on 06 Nov 2019, 10:20
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Bunuel
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)
Show more
P(5) + P(more than 5)=0.27
P(less than 5)=1-0.27=0.73
Not sufficient
From 2
P(5)+P(less than 5)=0.85
Not sufficient
From 1 and 2
P(5)=0.85-0.73
P(5)=0.12
C:)
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What is the probability that Lee will make exactly 5 errors on 05 May 2020, 11:05
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What is the probability that Lee will make exactly 5 errors on a certain typing test?

For those looking for Venn diagram solution.

(1) The probability that Lee will make 5 or more errors on the test is 0.27.
\(P_{>5} + P_{5} = 0.27\). Nothing about <5 errors is given.

INSUFFICIENT.

(2) The probability that Lee will make 5 or fewer errors on the test is 0.85.
\(P_{<5} + P_{5} = 0.85\). Nothing about >5 errors is given.

INSUFFICIENT.

Together 1 and 2. Refer diagram.
Attachment:
Probability 5 errors.png
Probability 5 errors.png [ 61.74 KiB | Viewed 25983 times ]
Adding both from statement 1 and 2
\(P_{>5} + P_{5} + P_{<5} + P_{5} = 0.27 +0.85\)
AND as \(P_{>5} + P_{5} + P_{<5} = 1\)

\(P_{5} = 1.12 - 1.0 = 0.12\)

SUFFICIENT.

Answer C.
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What is the probability that Lee will make exactly 5 errors on 23 Jun 2021, 05:35
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VeritasKarishma
Bunuel
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)


We know that

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors) ....... (I)

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

P (Exactly 5 errors) + P (More than 5 errors) = 0.27
Putting this in (I) we can get the value of P (Fewer than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

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

P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.85
Putting this in (I) we can get the value of P (More than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

Using both,
P (Exactly 5 errors) + P (More than 5 errors) + P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.27 + 0.85
P (Exactly 5 errors) + 1 = 1.12
P (Exactly 5 errors) = 0.12
Sufficient

Answer (C)
Show more

VeritasKarishma

how based on this info "What is the probability that Lee will make exactly 5 errors on a certain typing test?" you inferred that

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors)

i mean there could be other probabilities like (p) exactly 6, or any other probability of error ...

what type pf probability is this ? independent events or ....

i ask cause i associate such type of questions with this 1 = A+b-both+neither
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What is the probability that Lee will make exactly 5 errors on 23 Jun 2021, 21:34
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dave13
VeritasKarishma
Bunuel
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)


We know that

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors) ....... (I)

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

P (Exactly 5 errors) + P (More than 5 errors) = 0.27
Putting this in (I) we can get the value of P (Fewer than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

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

P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.85
Putting this in (I) we can get the value of P (More than 5 errors) but we cannot get the value of P (Exactly 5 errors).
Not sufficient

Using both,
P (Exactly 5 errors) + P (More than 5 errors) + P (Fewer than 5 errors) + P (Exactly 5 errors) = 0.27 + 0.85
P (Exactly 5 errors) + 1 = 1.12
P (Exactly 5 errors) = 0.12
Sufficient

Answer (C)

VeritasKarishma

how based on this info "What is the probability that Lee will make exactly 5 errors on a certain typing test?" you inferred that

1 = P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors)

i mean there could be other probabilities like (p) exactly 6, or any other probability of error ...

what type pf probability is this ? independent events or ....

i ask cause i associate such type of questions with this 1 = A+b-both+neither
Show more


dave13:
P (Fewer than 5 errors) + P (Exactly 5 errors) + P (More than 5 errors)
cover all possible cases and their probabilities.

P(exactly 6) is included in P(More than 5).
P(Exactly 7) is also included in P(More than 5).

In all possible cases, either there will fewer than 5 errors (so 0/1/2/3/4 errors) or there will be 5 errors or there will be more than 5 errors (so 6/7/8/9... errors).

These are mutually exclusive. If there are fewer than 5 errors, there cannot be 5 errors or more than 5 errors. Only one of these three events can take place at one time. Hence, both = 0 in all these cases. Also, neither = 0 because every case will fall in exactly one of these 3 probabilities.

Hence, the probabilities will simply be added.

1 = P(less than 5) + P(exactly 5) + P(more than 5)
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What is the probability that Lee will make exactly 5 errors on 02 Sep 2021, 10:01
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Correct option :C

to solve we need, making total [1]

A1. Error fewer than equal to 5 (statement 1)
A2. Error more than equal to 5 (statement 2)
A3. Exact Error 5 (non avaiable)

without wasting time, A, B, D are elimanated
left with C and E > can be solved or not be solved

P(A1) + P(A2) + P(A3) = 1 (Can be solved), makes C Winner
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What is the probability that Lee will make exactly 5 errors on 11 Oct 2025, 05:02
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