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-P(no letter in proper envelope)
1- (3C1 x 2)/9=1-2/3=1/3

Choose D

dtweah wrote:

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-P(no letter in proper envelope)
1- (3C1 x 2)/9=1-2/3=1/3

Choose D

Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see below
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.

maihuna wrote:

dtweah wrote:

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-P(no letter in proper envelope)
1- (3C1 x 2)/9=1-2/3=1/3

Choose D

Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see below
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 L1
E1 L2
E1 L3
E2 L1
E2 L2
E2 L3
E3 L1
E3 L2
E3 L3

dtweah wrote:

maihuna wrote:

dtweah wrote:

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-P(no letter in proper envelope)
1- (3C1 x 2)/9=1-2/3=1/3

Choose D

Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see below
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 L1
E1 L2
E1 L3
E2 L1
E2 L2
E2 L3
E3 L1
E3 L2
E3 L3

DTWEAH,
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

maihuna wrote:

dtweah wrote:

maihuna wrote:

dtweah wrote:

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-P(no letter in proper envelope)
1- (3C1 x 2)/9=1-2/3=1/3

Choose D

Nope Nope Big No, ans is definetly wrong, see there will not be 9 combination, see below
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 L1
E1 L2
E1 L3
E2 L1
E2 L2
E2 L3
E3 L1
E3 L2
E3 L3

DTWEAH,
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

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.

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....

maihuna wrote:Humm...your answer is incorrect, see yourself then where your fundamental principle of counting is putting you in trouble....

"An Envelope in a letter is different from an E's L's Combination." Why?

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?

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....

Attitude? What has that got to do with math. The FCP is a Law. I stand by it.

dtweah wrote:

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....

Attitude? What has that got to do with math. The FCP is a Law. I stand by it.

you mean answer is incorrect? what does it mean by your standing or sleeping or walking with FCP? Does it have any significance...?

You are correct about the 3 shirt - 3 pants combinations...there will be 3*3 = 9 combinations....but this problem is a little different...


The total possible combinations of letters and envelopes is 3! = 6

Think of it this way, the first letter (L1) has 3 choices (E1, E2, E3). So say we match L1 to E1. The second letter will have only 2 choices (E2, E3) left since E1 is already taken by L1. Say we match L2 to E2. Then the final letter has only 1 choice left (E3) as the other 2 are already taken.

Lets list the possible options...

E1L1 E2L2 E3L3 --> Atleast 1 letter in proper envelope
E1L1 E2L3 E3L2 --> Atleast 1 letter in proper envelope
E1L2 E2L1 E3L3 --> Atleast 1 letter in proper envelope
E1L2 E2L3 E3L1
E1L3 E2L1 E3L2
E1L3 E2L2 E3L1 --> Atleast 1 letter in proper envelope

Thats a total of 6 possible combinations. Can anyone list any additional combinations thats not already listed above?

Also, as you can, 4/6 have atleast 1 letter in proper envelope

Hence probability is 2/3.

Hope this helps.

Cheers.

dtweah wrote:

maihuna wrote:Humm...your answer is incorrect, see yourself then where your fundamental principle of counting is putting you in trouble....

"An Envelope in a letter is different from an E's L's Combination." Why?

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?

take letters A B and C
probability they wont fit in right envelope ..
A has two options ---- 2/3
B has 1 option ----1/2
C has 2 fit in the last one --1
multiply 2/3*1/2*1 = 1/3
therefore probability they fit = 1-1/3 = 2/3