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As the number of pens in the box after 3 withdrawal is 7, and defective pen remaining in the box is still 1. Probability of getting defective pen in 4th draw = 1/7

In this question, I don't think any reason to put back the pen back into the box again.
But if the pens are put back in the box, we get probability = 1/10, since the total number of pens in the box would be 10.

Answer to your query- "are the events independent? - NO

urvashis09
I think it might help to have the information whether the pen drawn is put back into the box. That is, are the events independent?
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As the number of pens in the box after 3 withdrawal is 7, and defective pen remaining in the box is still 1. Probability of getting defective pen in 4th draw = 1/7

In this question, I don't think any reason to put back the pen back into the box again.
But if the pens are put back in the box, we get probability = 1/10, since the total number of pens in the box would be 10.

Answer to your query- "are the events independent? - NO

urvashis09
I think it might help to have the information whether the pen drawn is put back into the box. That is, are the events independent?

Are we supposed to assume "without replacement" if nothing is mentioned in the question stem?
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You got it right...
If nothing is written, it means without replacement.


urvashis09
gmatbusters
As the number of pens in the box after 3 withdrawal is 7, and defective pen remaining in the box is still 1. Probability of getting defective pen in 4th draw = 1/7

In this question, I don't think any reason to put back the pen back into the box again.
But if the pens are put back in the box, we get probability = 1/10, since the total number of pens in the box would be 10.

Answer to your query- "are the events independent? - NO

urvashis09
I think it might help to have the information whether the pen drawn is put back into the box. That is, are the events independent?

Are we supposed to assume "without replacement" if nothing is mentioned in the question stem?
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Can someone please give a mathematical explanation of the answer, with steps?
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Bunuel
Only one of ten pens in a box is defective. Three pens are randomly drawn from the box, one at a time. If none of those three pens is defective, what is the probability that the next pen randomly drawn is defective?

(A) 1/10
(B) 1/7
(C) 7/10
(D) 6/7
(E) 9/10

total pens = 10
so first three pens are taken out
left with 7 pens
since only 1 pen is defective so P = 1/7
IMO B
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Bunuel
Only one of ten pens in a box is defective. Three pens are randomly drawn from the box, one at a time. If none of those three pens is defective, what is the probability that the next pen randomly drawn is defective?

(A) 1/10
(B) 1/7
(C) 7/10
(D) 6/7
(E) 9/10


Since we have 7 pens left after three non-defective pens are drawn, and one of the 7 pens is defective, the probability that the 4th pen drawn is defective is 1/7.

Answer: B
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