arjunsridhar84 wrote:
Hi Bunuel - i didnt understand how yangsta arrived at 1*2*1*1 for all incorrect or for that matter 1*1*1*1 for one incorrect and so on... i didnt quite follow this logic.
kindly explain
Counting 1 correct: 4C1=4, choosing which letter will be put in correct envelope. (Let's assume A letter was put in correct envelope).
Then 3
letters (BCD) and 3
envelopes (BCD) are left. # of ways to put them incorrectly is 2:
Envelopes: B-C-DLetters: C-D-B
OR: D-B-C
So total # of ways one letter will be put into the envelope with its correct address is: 4C1*2=8. As total # of ways to distribute 4 letter in 4
envelopes is 4!=24, probability P(C=1)=8/24.
Counting 2 correct: 4C2=6, choosing which 2 letter will be put in correct envelope. (Let's assume A and B letter were put in correct
envelopes).
Then 2
letters (CD) and 2
envelopes (CD) are left. # of ways to put them incorrectly is 1:
Envelopes: C-DLetters: D-C
So total # of ways two
letters will be put into the envelope with its correct address is: 4C2*1=6. As total # of ways to distribute 4 letter in 4
envelopes is 4!=24, probability P(C=2)=6/24.
Counting 3 correct: if three
letters will be put in correct
envelopes, then the fourth one also gets its correct envelope. So there is no way we can put exactly 3
letters in correct
envelopes --> P(C=3)=0
Counting 4 correct: there are 4!=24 ways to distribute 4
letters in 4
envelopes and obviously only one is when all
letters get their correct
envelopes. P(C=4)=1/24.
Counting all incorrect, or 0 correct:P(all incorrect)=1-(1 correct)-(2 correct)-(3 correct)-(4 correct)=1-8/24-6/24-0-1/24=9/24.
ichha148 wrote:
A. That no letter will be put into the envelope with its correct address?
9/24
B. That all
letters will be put into the envelope with its correct address?
1/24
I understand that correct answer is 9/24 , however my question is should not the B. That all
letters will be put into the envelope with its correct address? is opposite of no letter will be put into the envelope with its correct address?
So , should not the result be 1-1/24 = 23/24
Can some one please explain me why this is not 23/24 and when 23/24 is applicable
Opposite of "all
letters in correct
envelopes" (= 1-1/24 = 23/24), is: "at least one in wrong envelope" = "3 correct" (=0) + "2 correct" (=6/24) + "1 correct" (=8/24) + "0 correct, option A" (=9/24) --> \(1-\frac{1}{24}=\frac{23}{24}=0+\frac{6}{24}+\frac{8}{24}+\frac{9}{24}=\frac{23}{24}\).
Hope it's clear.