1-P(no letter in proper envelope)maihuna wrote:Three letters are dictated to three persons and an evelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
2/9
3/9
4/9
1/3
2/3
Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see belowdtweah wrote:1-P(no letter in proper envelope)maihuna wrote:Three letters are dictated to three persons and an evelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
2/9
3/9
4/9
1/3
2/3
1- (3C1 x 2)/9=1-2/3=1/3
Choose D
E1 L1maihuna wrote:Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see belowdtweah wrote:1-P(no letter in proper envelope)maihuna wrote:Three letters are dictated to three persons and an evelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
2/9
3/9
4/9
1/3
2/3
1- (3C1 x 2)/9=1-2/3=1/3
Choose D
1 st letter have 3 option
2nd letter have 2 option
3rd letter have 1 option
so total event space is only 3*2*1 = 6 not 9 as it will be when enumerating E1,E2,E3 with L1,L2,L3.
DTWEAH,dtweah wrote:E1 L1maihuna wrote:Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see belowdtweah wrote:1-P(no letter in proper envelope)maihuna wrote:Three letters are dictated to three persons and an evelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
2/9
3/9
4/9
1/3
2/3
1- (3C1 x 2)/9=1-2/3=1/3
Choose D
1 st letter have 3 option
2nd letter have 2 option
3rd letter have 1 option
so total event space is only 3*2*1 = 6 not 9 as it will be when enumerating E1,E2,E3 with L1,L2,L3.
E1 L2
E1 L3
E2 L1
E2 L2
E2 L3
E3 L1
E3 L2
E3 L3
By Fundamental Counting Principle If you have 3 shirts and 3 trousers, the number of outfits you can wear is 9. Three Letters and Three envelopes are no different than 3 trousers and and 3 shirts. By your reasoning if I pick one shirt, I am left with only 2 shirts. Total choices are not calculated that way. Wrong. You have 3 choices of Evelopes and letters for your first choice. If you match evelope to letter put it back and examine the other choices. Why are you leaving letters in envelopes to examine choices? We are interested in ALL possibilites. Do the experiment. You will have 9 choices as outlined.maihuna wrote:DTWEAH,dtweah wrote:E1 L1maihuna wrote:Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see belowdtweah wrote:1-P(no letter in proper envelope)maihuna wrote:Three letters are dictated to three persons and an evelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
2/9
3/9
4/9
1/3
2/3
1- (3C1 x 2)/9=1-2/3=1/3
Choose D
1 st letter have 3 option
2nd letter have 2 option
3rd letter have 1 option
so total event space is only 3*2*1 = 6 not 9 as it will be when enumerating E1,E2,E3 with L1,L2,L3.
E1 L2
E1 L3
E2 L1
E2 L2
E2 L3
E3 L1
E3 L2
E3 L3
Thats what I was pointing, an envelope in a letter is different from an E's L's combination, pick three letter and three envelope and experiment, how many arrangements will be possible? First letter has three envelope to go, once you do that you are left with two envelope and two letter, so an letter have 2 option, for last one there is only one option, so combining one gets onbly six option.
Out of these 6 in following way one or more letter may go to correct place:
(E1, L1) (E2, L3) (E3, L2)
similarly other two may be choosen, i.e. E2,L2 and E3 L3 in one way only
so for one letter in correct place there are 3 ways
For two letter, if two letter are in correct place third will be automatically in correct place, so only one way of getting it right for more than one..
so a total of 3+1 = 4 ways of getting it right
Reqd Prob = 4/6 = 2/3
"An Envelope in a letter is different from an E's L's Combination." Why?maihuna wrote:Humm...your answer is incorrect, see yourself then where your fundamental principle of counting is putting you in trouble....
Attitude? What has that got to do with math. The FCP is a Law. I stand by it.maihuna wrote:Humm...your answer is incorrect, I am seeing where you are making mistake, but I will not point that out, as your attitude is very un-accepting sort of....see yourself then where your fundamental principle of counting is putting you in trouble....
you mean answer is incorrect? what does it mean by your standing or sleeping or walking with FCP? Does it have any significance...?dtweah wrote:Attitude? What has that got to do with math. The FCP is a Law. I stand by it.maihuna wrote:Humm...your answer is incorrect, I am seeing where you are making mistake, but I will not point that out, as your attitude is very un-accepting sort of....see yourself then where your fundamental principle of counting is putting you in trouble....
dtweah wrote:"An Envelope in a letter is different from an E's L's Combination." Why?maihuna wrote:Humm...your answer is incorrect, see yourself then where your fundamental principle of counting is putting you in trouble....
E1 E2 E3 L1 L2 L3
L1 can go in E1 , that is one possibility
L1 can go into E2, that is another possibility. What prevents this in the wording of the problem?
L1 can go into L3, a third possibility. What prevents this?
