Latest TCS Aptitude Question SOLUTION: A father purchases dress for his three daughter.The dresses are of same color,but of different size.The dress is kept in dark room.What is the probability that all the three will n

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A father purchases dress for his three daughter.The dresses are of same color,but of different size.The dress is kept in dark room.What is the probability that all the three will not choose their own dress..

Asked In TCS RIYAM TAHER (9 years ago)

Unsolved Read Solution (8)

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Pranoy Bej  (9 years ago)

ANSWER==> 1/3
the no of ways the dress can be arranged= 3! = 6
1St girl can select her dress wrongly(probabilty) in 2/3 way
so there is 2 dress left
2nd girl can choose her dress wrongly(probability)= 1/2
last girl have to choose the last dress left in only 1 way.

SO probabilty of Three Girl choose the dress wrongly P(E)= 2/3*1/2*1= 2/6= 1/3

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Niveditha Ravichandran  (9 years ago)

let the girls be x,y and z
x y z
---------
z x y
y z x

there are 2 chances where the girls do not choose their own dress
total chance is 3!= 6

so probability = 2/6= 1/3

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

abdul quadir khan  (9 years ago)

26/27
As choosing the dress have sample of 3^2 and only one way to choose the correct dress
1-(1/27)=26/27

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Naveen Kodali  (9 years ago)

Assume Daughters be X Y & Z
their corresponding dress size be X' Y' & Z'
so the total probability for picking odd one is
X-Y', X-Z', Y-X', Y-Z', Z-X' and Z-Y' so 6 ways.
all the possible ways to pick up a dress are
X-X', X-Y', X-Z', Y-X', Y-Y', Y-Z', Z-X', Z-Y' and Z-Z' so 9 ways.
the probability here is
6/9 ==2/3
Answer is 2/3

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Raghav gupta  (9 years ago)

they have asked in the question that what is the prob that all of the three chooses there dress wrongly only two cases are in favour out of 6 that is 2/6=1/3 and this is the case of dearrangements so dont subtract that dey all will choose there dress right from total possibility as it will also give the possibility that one will select wrong or two will select wrong

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MUNISH YADAV  (9 years ago)

Total ways of selecting the dresses is 3!
Probability of selecting the correct match is 1/3!
Thus, probability of never selecting the correct match is 1-1/3!=1-1/6=5/6

Is this solution Helpfull? Yes | No (7)  

Shahnawazalam  (9 years ago)

A did 4% work in one day...they both did 20/3 =6.66% of work in one day so B did 2.66% of work in one day...therefore b tooks 100/2.66=37.5days

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abhyuday pal  (9 years ago)

E = All three do not choose the correct dress
F = All three choose their dresses correctly
Obviously, P(E)= 1-P(F)
Now ther's only 1 way for each one of them to choose the dress correctly
Therefore, P(F)= 1/(3!) = 1/6
and hence, P(E)= 1- (1/6) = 5/6

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