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One of 5 pens is faulty. They are tested one at a time, what is the pr

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Bunuel
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One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   02 Mar 2021, 03:15
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One of 5 pens is faulty. They are tested one at a time, what is the probability that the faulty pen is the 3rd one tested ?

A. 12/125
B. 1/5
C. 2/5
D. 3/5
E. 4/5
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B
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CrackverbalGMAT
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One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   Updated on: 02 Mar 2021, 19:32
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P(First Good Pen) = 4/5

P(Second good pen) = 3/4

P(3rd pen is faulty) = 1/3

The required probability = 4/5 * 3/4 * 1/3 = 1/5


Option B

Arun Kumar

Originally posted by CrackverbalGMAT on 02 Mar 2021, 05:32.
Last edited by CrackverbalGMAT on 02 Mar 2021, 19:32, edited 1 time in total.
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jrk23
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   02 Mar 2021, 10:45
CrackVerbalGMAT wrote:
Hi Bunuel, could you please check the answer Options? IMO, the answer should be 16/125

P(Good Pen) = 4/5

P(3rd pen is faulty) = 4/5 * 4/5 * 1/5 = 16/125


Arun Kumar


Can't we say that if third pen was defected than first 2 is also defective. thus, probability = 4/5 * 3/4 * 1/3 (I mean after checking 1st pen 4 left so probabilty of being defective after testing first is 3/4) and same with third.

In above case how question would be written... if p= 4/5 * 3/4 * 1/3

Thanks
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CrackverbalGMAT
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   02 Mar 2021, 17:30
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jrk23 wrote:
CrackVerbalGMAT wrote:
Hi Bunuel, could you please check the answer Options? IMO, the answer should be 16/125

P(Good Pen) = 4/5

P(3rd pen is faulty) = 4/5 * 4/5 * 1/5 = 16/125


Arun Kumar


Can't we say that if third pen was defected than first 2 is also defective. thus, probability = 4/5 * 3/4 * 1/3 (I mean after checking 1st pen 4 left so probabilty of being defective after testing first is 3/4) and same with third.

In above case how question would be written... if p= 4/5 * 3/4 * 1/3

Thanks


Hi. In that case, the probability of getting a faulty pen will always be 1/5, whether it comes 1st, 2nd 3rd 4th or 5th
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chetan2u
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   02 Mar 2021, 19:00
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CrackVerbalGMAT wrote:
Hi Bunuel, could you please check the answer Options? IMO, the answer should be 16/125

P(Good Pen) = 4/5

P(3rd pen is faulty) = 4/5 * 4/5 * 1/5 = 16/125


Arun Kumar



The answer will be 1/5.
You have made an error in 4/5 * 4/5 *1/5. It should be 4/5 *3/4 *1/3=1/5
The probability will keep changing and will not remain constant. This would be the case if we were replacing the pen after testing.
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   02 Mar 2021, 19:04
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Bunuel wrote:
One of 5 pens is faulty. They are tested one at a time, what is the probability that the faulty pen is the 3rd one tested ?

A. 12/125
B. 1/5
C. 2/5
D. 3/5
E. 4/5



Logically
Each pen should have the same probability of being defective. Thus 1/5 is the probability

Solution
The probability of first pen being good =4/5
The probability of the 2nd pen being good =3/4, as there are only 4 pens left now.
The probability of the third pen being defective =1/3, as there are only 3 pens left and one is defective
P=\(\frac{4}{5}*\frac{3}{4}*\frac{1}{3}=\frac{1}{5}\)

B
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   02 Mar 2021, 19:34
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chetan2u wrote:
CrackVerbalGMAT wrote:
Hi Bunuel, could you please check the answer Options? IMO, the answer should be 16/125

P(Good Pen) = 4/5

P(3rd pen is faulty) = 4/5 * 4/5 * 1/5 = 16/125


Arun Kumar



The answer will be 1/5.
You have made an error in 4/5 * 4/5 *1/5. It should be 4/5 *3/4 *1/3=1/5
The probability will keep changing and will not remain constant. This would be the case if we were replacing the pen after testing.


Hi chetan2u. Basically, irrespective when the faulty pen comes (whether 1st, 2nd, 3rd, 4th or 5th), the probability will always be 1/5, right?
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   04 Mar 2021, 13:35
CrackVerbalGMAT wrote:
P(Good Pen) = 4/5

P(3rd pen is faulty) = 4/5 * 4/5 * 1/5 = 16/125


Arun Kumar


Can't we say that if third pen was defected than first 2 is also defective. thus, probability = 4/5 * 3/4 * 1/3 (I mean after checking 1st pen 4 left so probabilty of being defective after testing first is 3/4) and same with third.

In above case how question would be written... if p= 4/5 * 3/4 * 1/3

Thanks[/quote]

That mean my thought was right.

Answer is 1/5 not 16/25...... m i right?

Hi. In that case, the probability of getting a faulty pen will always be 1/5, whether it comes 1st, 2nd 3rd 4th or 5th[/quote]
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink] New post   10 Mar 2021, 04:21
We go with the constraints in the question.

First pick - non defective (1 out 4 non defective) - Probability= 4/5
Second pick - again non defective (1 out of 3 non defective left, as one got picked in the first pick) - Probability = 3/4
Third pick - defective pick (1 out of 3 remaining, as the other 2 are free from defects) - Probability = 1/3

So for this sequential scenario, we just multiply the individual event probabilities
4/5 * 3/4 * 1/3
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Re: One of 5 pens is faulty. They are tested one at a time, what is the pr [#permalink]
10 Mar 2021, 04:21
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