Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability?
A.
The probability that 1st letter will not be put in correct address is 3/4
The probability that 2nd letter will not be put in correct address is 2/3
The probability that 3rd letter will not be put in correct address is ½
The probability that 4th letter will not be put in correct address is 1
Combine probability is ¾*2/3*1/2*1 = ¼
B.
The probability that 1st letter will be put in the correct address is ¼
The probability that 2nd letter will be put in the correct address is 1/3
The probability that 3rd letter will be put in the correct address is ½
The probability that 4th letter will be put in the correct address is 1
Combine probability = ¼*1/3*1/2*1 = 1/24
C.
The probability that 1st letter in correct address is ¼
The probability that 2nd letter not in correct address = as there are 2 wrong and 1 right address 2/3
The probability that 3rd letter not in correct address = as there are 1right and 1 wrong address, hence ½
The probability that last letter in wrong address is 1
The combine probability of 1st letter in right address and remaining letters in wrong is = ¼*2/3*1/2 = 1/12
As any one letter can be in right address we have to multiply the above probability by 4
1/12 * 4 = 1/3
Hence, probability that exactly one letter in right address is 1/3
D.
Probability that the first letter in correct address is ¼
Probability that the second letter in correct address is 1/3
Probability that third letter in wrong address is ½
Probability that last letter in wrong address is 1
Combine probability = 1/24
There can be 6 combinations of 2 right and 2 wrong addresses
Hence probability that exactly 2 letters in right address is 1/4
E.
Exactly 3 letters will be put in correct address means all in correct address.
If 3 letters are in correct address ,then obviously last letter will go in correct address
probability is 0
F.
P ( more than one letter in right envelope correct ) = 1- P( exactly 1 letter in right envelope correct)
= 1- 1/3 = 2/3
G.
P ( more than 2 letters in right envelope correct ) = 1- p (exactly 2 letters in right envelope )
= 1- 1/8 = 7/8
Looking at the answer by other members,I figure out I made lot of mistakes........ I am not able to figure out where I am going wrong... Please help me out with my approach
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-------Analyze why option A in SC wrong-------