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

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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(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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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

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 .
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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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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

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

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

Posted from my mobile device

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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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:)

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.
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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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

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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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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