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In an interview, the following question was asked:

A (single) box contains $20$ pens. $8$ pens are of type $A$, and the others are of type ‌$B$. We then randomly choose a pen, and discard it from the box (regardless of type). If we then choose another pen from the box, what is the probability that this pen is of type $A$?

I'm having trouble coming up with a solution to this, how would you compute this?

bilibraker
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4 Answers4

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How can I select a pen randomly? Supposing that it is not possible to detect which pen is which merely by feeling the pens (because the difference is only their color, for example), I can shake up the contents of the box, perhaps reach in and stir them with my hand, then finally grasp a pen and pull it out.

Suppose that while I am stirring the pens I grasp one and then put my fingers on a second pen; I then carefully grasp the second pen while dropping the first pen so that I make sure the first pen is not the one I am holding, and I pull the second pen from the box.

With respect to the probability that I finally choose a type A pen, how is this procedure different from any other method of stirring the pens and finally pulling one out?

With respect to the probability that I finally choose a type A pen, how is this procedure different from putting the first pen grasped in an empty corner of the box and then grabbing the second pen, ensuring that I do not take the first pen?

With respect to the probability that I finally choose a type A pen, how is this procedure different from removing the first pen without looking and putting it in some other place outside the box where I still do not see it, and then grabbing the second pen?

I think there is no need here for any calculation more involved than dividing one number by another.

David K
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    you means another answer is wrong? –  Mar 14 '15 at 13:29
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    No, I mean the probability is the same as picking up a pen of type A in the first place. Same result as the other answer, but without the calculation. – David K Mar 14 '15 at 13:39
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    +1 nice point. It then generalizes to having $100$ pens and eliminating the first 37. Exact computation become cumbersome :-) – Ant Mar 14 '15 at 16:26
  • Awesome! That's a fantastic way to think about the question. – A Googler Apr 21 '15 at 10:18
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You have two scenarios to sum together : First pen is type B or first pen is type A. The probabilities for these are $\frac{12}{20}\times\frac{8}{19}$ and $\frac{8}{20}\times\frac{7}{19}$, which sums to the answer $\frac{8}{20}$, which is the same as the probability of picking up pen A the first time round.

Alijah Ahmed
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$P(A)=8/20=0.4, P(B)=0.6$

We have to compute the conditional probability

$P(A|A\cup B)=P(A\cap(A\cup B))/(P(A\cup B))=P(A)/(P(A)+P(B))=P(A)=0.4$

Here is a Python simulation:

import numpy as np
from random import randint
rez=[]   
N=1000
for k in range(N):
    u=np.random.random()
    if  u<0.4: # an A-type is drawn
        x=randint(1,19)#second draw
        if x<8: rez.append(1)
    else: # a B-type is drawn
        x=randint(1,19)#second draw
        if  x<=8: rez.append(1)

 print 1.0*sum(rez)/N
petem
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Hint: either the first pen is of type A or of type B. Can you compute the probability that the second pen is of type A in each case ?

bela83
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