Latest TCS Aptitude Question SOLUTION: there are 5 envelope and 5 letters placed at random. what is the probability that no single letter is placed in the right envelope

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there are 5 envelope and 5 letters placed at random. what is the probability that no single letter is placed in the right envelope

Asked In TCS nidhi gupta (9 years ago)

Solved Pranoy Bej Read Solution (8)

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Read Solution (8) :

sahitya  (9 years ago)

probability that no single letter is placed in the right envelop is = 1- (placing all the letters in the right envelop)
i.e 5 letters will go in their right envelops in only 1 way, and total cases is 5!
so, 1-(1/5!) = 119/120

Is this solution Helpfull? Yes (14) | No (1)  

Pranoy Bej  (9 years ago)

If we let A be the event that letter A is in ist correct envelope and
similarly B is the event that letter B is in its correct envelope,
then

P(A) = 1/5 and

P(A and B) = 1/5 * 1/4

Now use the inclusion-exclusion principle to get probability that A or
B or C .... or E are correctly placed.

P(A or B or C .... or E) = P(A) + P(B) + P(C) + P(D) + P(E)
- P(A and B) - P(B and C) -....
+ P(A and B and C) + P(B and C and D) + ....
- P(A and B and C and D) - P(...) -......
+ P(A and B and C and D and E)

= 5*(1/5)
- 5C2*(1/5)(1/4)
+ 5C3*(1/5)(1/4)(1/3)
- 5C4*(1/5)(1/4)(1/3)(1/2)
+ (1/5)(1/4)(1/3)(1/2)(1/1)

5*4 1 5*4*3 1 5*4*3*2 1 1
= 1 - --- * --- + ----- * ----- - --------- * ------- + ---------
1*2 5*4 1*2*3 5*4*3 1*2*3*4 5*4*3*2 5*4*3*2*1

= 1 - 1/2! + 1/3! - 1/4! + 1/5! = 19/30

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anshul sharma  (9 years ago)

p of placing them in right envelopes is 1/5! hence for not placing them is 1-1/5!=119/120

Is this solution Helpfull? Yes (4) | No (2)  

radhika mittal  (9 years ago)

case of total disagreement so formula is n!(1/2!-1/3!+1/4!........+1/n!)
so 5!(1/2!-1/3!+1/4!-1/5!)=44
so p=44/5!=44/120=11/30.

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nishant samrendra  (9 years ago)

only 15 cases(1+2+3+4+5) are there in which each letter will go into correct envelope, while total 5^5 cases are there to place the letters into envelope....
ans is : 5^5 - 15.

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shanmathi  (9 years ago)

anyone pls explain this prob clearly.

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yogesh ahuja  (9 years ago)

here it is said no single letter should be at correct place
so 1stly chances of getting 1 letter at correct place=5
2ndly chances of getting 2 letters at correct place=5c2=10
3rdly chances of getting 3 letters at correct place=5c3=10
lastly chances of getting 5 letters at correct place=1
note that 4 letters cant be at rt place bcoz it will ultimately lead all letters at correct place
sum=5+10+10+1=26
chance of no letter at correct place=120-26=94
p=94/120

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ravi prakash  (9 years ago)

question is all of them are wrong not only one of them

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