Q. Tanya prepared 4 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 into 4 envelopes at random, what is the probability that only 1 letter will be put its envelope with correct address?
One of the answer explanations which came was -
C=correct
I=incorrect
so lets say the order of letters is CIII
now first letter can be put in correct envelope by 1/4
now second letter can be put in incorrect envelope by 2/3
now third letter can be put in incorrect envelope by 1/2
now fourth letter can be put in incorrect envelope by 1
1/4+2/3+1/2+1=1/12
now CIII can be rearranged in 4C1=4 ways
probability=4*1/12=1/3 way
My Doubt
let us say the letters and envelopes are L1,E1,L2,E2,L3,E3,L4,E4
now let us take CIII....
Prob of correct envelope for L1 = 1/4
Prob of incorrect envelope for L2 = 2/3....(here L2 can pick E3 or E4)
If L2 picks E3, then prob of incorrect envelope for L3 will be 1 (only E2 and E4 left) and similarly prob of incorrect envelope for L4 will be 1 too...
am i thinking correct??
One of the answer explanations which came was -
C=correct
I=incorrect
so lets say the order of letters is CIII
now first letter can be put in correct envelope by 1/4
now second letter can be put in incorrect envelope by 2/3
now third letter can be put in incorrect envelope by 1/2
now fourth letter can be put in incorrect envelope by 1
1/4+2/3+1/2+1=1/12
now CIII can be rearranged in 4C1=4 ways
probability=4*1/12=1/3 way
My Doubt
let us say the letters and envelopes are L1,E1,L2,E2,L3,E3,L4,E4
now let us take CIII....
Prob of correct envelope for L1 = 1/4
Prob of incorrect envelope for L2 = 2/3....(here L2 can pick E3 or E4)
If L2 picks E3, then prob of incorrect envelope for L3 will be 1 (only E2 and E4 left) and similarly prob of incorrect envelope for L4 will be 1 too...
am i thinking correct??
