Question

Pratibha’s desk has 8 drawers. When she receives a paper, she usually chooses a drawer at random to put it in. However, 2 out of 10 times she forgets to put the paper away, and it gets lost.
The probability that a paper will get lost is 2/10 , or 1/5 .
- What is the probability that a paper will be put into a drawer?
- If all drawers are equally likely to be chosen, what is the probability that a paper will be put in drawer 3?
When Pratibha needs a document, she looks first in drawer 1 and then checks each drawer in order until the paper is found or until she has looked in all the drawers.
1. If Pratibha checked drawer 1 and didn’t find the paper she was looking for, what is the probability that the paper will be found in one of the remaining 7 drawers?
2. If Pratibha checked drawers 1, 2 and 3, and didn’t find the paper she was looking for, what is the probability that the paper will be found in one of the remaining 5 drawers?
3. If Pratibha checked drawers 1–7 and didn’t find the paper she was looking for, what is the probability that the paper will be found in the last drawer?

Answer 1

(i) the probability that a paper will be put into a drawer is 8/10 or 4/5.
(ii) the probability that a paper will be put in drawer 3 is 1/8.
(iii) the probability that the paper will be found in one of the remaining 7 drawers is 1/7.
(iv) the probability that the paper will be found in one of the remaining 5 drawers is 1/5.
(v) the probability that the paper will be found in the last drawer is 1/1 or 1 i.e. she will definitely find that paper in the last drawer.

Answer 2

   Probability that the paper piece is lost = 1/5

   Probability that the paper is not lost or put into some drawer = 1 - 1/5 = 4/5

   Probability that the paper is put in to a particular drawer = 4/5 * 1/8 = 1/10.
So probability that the paper is put in to 3rd drawer = 1/10

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1.        P(A | B)  =  P(A Π B) / P(B)
   Probability of event A  given that event B occurred is the probability of A and B occurring together divided by probability of B.

   A =  paper is found in a drawer of the 7 remaining drawers.
   B =  paper is NOT found in the 1st drawer.

   P(B) = paper is not put in any drawer + if it is put in a drawer then, paper is put in other 7 drawers
           = 1/5 +  4/5 * 7/8    = 9/10
   P(A) = 4/5 * 7/8 = 7/10
   P(A U B) = 1/5 + 7/10 = forgotten + in 7 drawers
                  = 9/10

   P(A Π B) = P(A) + P(B) - P(A U B) = P(A)
                 = 7/10  + 9/10 - 9/10  =  7/10

   P(A | B) = [ 7/10 ]  /  [ 9/10 ]  = 7/9
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2.  B' = paper is found in drawers 1 or 2 or 3 = 3/10
     P(B)  =  paper is NOT found in drawers 1, 2 and 3 (any):
               = 1/5 + 4/5 * 5/8 = 7/10
     P(A)  =  paper is in drawers any of 4,5,6,7, or 8 = 5/10 = 1/2
     P(A U B) = 2/5 + 1/2 = 9/10
     P(A Π B ) = P(A) + P(B) - P(AUB)
                    = 1/2 + 7/10 - 9/10 = 3/10
 
   P(A | B) =  [ 3/10 ]  /  [7/10]
                 = 3/7
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3.     [1/10]  / [3/10]
   this is done by using another method.
 
     The paper is not put in the first 7 drawers.   So the paper is forgotten to be put in to a drawer  OR put in the 8th drawer.  Both are exclusive events.
A = paper is put in 8th drawer
B = paper is forgotten to be put in a drawer
P(A Π B) = 0      and    there are only two independent and exclusive events.

   P(A | paper is not in 7 drawers) = P(A) / [ P(A) +P(B) ]
                     = [1/10] / [1/10 + 1/5]         = 1/3

  using the method like the above two parts 1 and 2:

  B = paper is NOT in 7 drawers = 1 - 7/10 = 3/10         {OR,  1/5 + 1/10 = 3/10  }
  A = paper is in the 8th drawer = 1/10

  P(A U B) = 1/5 + 1/10 = 3/10
  P(A Π B) = 3/10 + 1/10 - 3/10 =  1/10

   P(A | B) = P(A Π B) / P(B) = [1/10]  / [3/10] = 1/3