Latest TCS Aptitude Question SOLUTION: A father purchased dress for his 3 daughters. The dresses are of same color but diff size and they are kept in dark room. what is probability that all the 3 will not choose there

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A father purchased dress for his 3 daughters. The dresses are of same color but diff size and they are kept in dark room. what is probability that all the 3 will not choose there own dress?

Asked In TCS sahana (10 years ago)

Solved k.m Read Solution (13)

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Read Solution (13) :

k.m  (10 years ago)

probability is 1/3 because
let 1st girl come and she choose wrong dress so probability of that girl to choose wrong dress out of 3 is =2/3.
now 2nd girl come nd she choose wrong dress so probability of that girl to choose wrong dress out of 2 is 1/2.
now for 3rd girl probability is 1 to choose wrong dress.
so probability tht all the 3 wil not choose der own dress is =2/3*1/2*1=1/3.
ans is 1/3.

Is this solution Helpfull? Yes (153) | No (6)  

Nirmalram  (10 years ago)

p(not choose own dress)=1-p(choosing own dress)

p(choose own dress)=(1/3)(1/2)(1)
=1/6

ans: p(not choose own dress=1-(1/6)
=5/6

Is this solution Helpfull? Yes (31) | No (37)  

vinay kumar  (10 years ago)

Total selections=3!=6

Proper Selections:
A B C
A C B
C B A
B A C (ie. total 4)

Improper selection: B C A , C A B (ie. total 2)
Probability for improper selection= 2/6 = 1/3

Is this solution Helpfull? Yes (25) | No (4)  

ANKUR SINGH  (10 years ago)

( 2/3 * 1/2 * 1 )= 1/3 this is the probability that they will not choose their own dress

Is this solution Helpfull? Yes (13) | No (5)  

Prakash Singh  (10 years ago)

(2/3)*(1/2)*1=1/3

Is this solution Helpfull? Yes (4) | No (1)  

akshat sinha  (8 years ago)

correct answer is 1/3

Method 1)
3 daughters (correct pairs daughter no. 1-a 2-b 3-c)
probability that daughter 1 will choose correct dress (i.e a) is 1/3
probability that daughter 1 will not choose a is 1- 1/3 = 2/3.
remaining dresses 2 out of which only 1 pair is correct (as first daughter picked someone else's dress)
so if either 2 or 3 pick incorrect dress with probability of 1/2 last pair is guaranteed wrong
total probability = 2/3 * 1/2 * 1 = 1/3



method 2)
let A`s dress named A, B dress named B, C`s dress named C.
Then
Total Combinations are 3!=6

A B C
A C B
C B A
B A C
B C A
C A B

Now, We have to find a condition where
A does not choose her dress,
B does not choose her dress and
C does not choose her dress.

In first four condition,
either A or B or C choose either of its right choice
like in ACB, A choose right choice... and so on
but in last two cond.
B C A
C A B
None of them chooses right choice
So fav outcome=2
and total outcome=3!=6
prob-2/6=1/3

Is this solution Helpfull? Yes (3) | No   

ankit chauhan  (10 years ago)

for 1st one= 1/3 second one 1/2 third one=1 total=1/6

Is this solution Helpfull? Yes (2) | No (3)  

ankita gajanan kulkarni  (10 years ago)

ans is 5/6

Is this solution Helpfull? Yes (2) | No (5)  

ankita gajanan kulkarni  (10 years ago)

ans is 5/6

Is this solution Helpfull? Yes (2) | No (5)  

arvind  (10 years ago)

vinay kumar's ans is correct and explanation too
ans is 1/3

Is this solution Helpfull? Yes (2) | No   

aditi patel  (10 years ago)

Formula for case of de-arrangements = n!(1/2! - 1/3! - ... - 1/n!)
hence, no. of ways dt none of them chooses there own dress = 3!(1/2!-1/3!)=2

P=2/3!=2/6=1/3

Is this solution Helpfull? Yes (2) | No (1)  

Abhijit Das  (10 years ago)

5/6
as 1/6 is the probabilty of choosing thier specific dress ,so 1-1/6 is d ans

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

Vijendra Kumar Rana  (10 years ago)

1/9+1/9+1/9=1/3

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