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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?

(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.

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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

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 ============================ 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