Latest TCS Aptitude Question SOLUTION: My name is PREET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unc

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(#M40010529) TCS Numerical Ability question Keep an EYE

My name is PREET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
a) 5 %
b) 10%
c) 20%
d) 25%
e) 12.5%

Asked In TCS vanithamateti (11 years ago)

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Aman  (11 years ago)

10%
son can interchange a pair in 10 combination out of which only one keps the name unchanged.
10 combinations for change are
PR
PE
PE
PT
RE
RE
RT
EE
ET
ET

so the probability that despite this interchange, the name remains unchanged = 1/10 = 10%

Is this solution Helpfull? Yes (62) | No (6)  

shweta  (11 years ago)

no of ways in which name remain same is 1 by intrchng e,e.noof ways of intrchng 2 letters is 5C2=10.thre4 probality is 1/10 .so ans is 10%.

Is this solution Helpfull? Yes (39) | No (3)  

lala  (11 years ago)

preet
no of sample space in which change can be happen is = {E,E}..........i.e number of possible set = 1
and total set is = 5 (because 5 letters are there)
so probability = 1/5
% probability = 1/5*100
equals 20

Is this solution Helpfull? Yes (10) | No (24)  

Yash Ajmera  (11 years ago)

Ans. 25%
According to Q "A pair of letters was inter changed. So possible pairs in PREET are
1.PR
2.RE
3.EE
4.ET
(Rest combinations did not make a pair in PREET like PE, PT, etc)
Total 4 out of which only "EE" is the one required
so probability = 1/4 or 25%.

Is this solution Helpfull? Yes (6) | No (7)  

RIMAN  (11 years ago)

ANS : c) 20%
Here, by interchanging the letters we can construct
5!/2(divided by 2 because of same letter 'E' is used twice here )
=60 different names.
Now his son type his name by interchanging a pair of letter.
So we can get
(5-1)C2 (because here E letter is used twice)
=4C2 = 12 different pair.
Now our required probability is 12/60 =1/5 or (1/5 * 100%) = 20%

Is this solution Helpfull? Yes (4) | No (17)  

varma  (11 years ago)

preet
no of sample space in which change can be happen is = {E,E}..........i.e number of possible set = 1
and total set is = 5 (because 5 letters are there)
so probability = 1/5
% probability = 1/5*100
equals 20

Is this solution Helpfull? Yes (4) | No (9)  

nagesh  (11 years ago)

ther are 5 chance of interchanging pair of letters are (P.R),(R,E),(E,E),(E,T),(T,P), but if you interchange (E,E) there is no change in name so probability of same name is 1/5 so ans is 20%

Is this solution Helpfull? Yes (3) | No (14)  

parul mehta  (10 years ago)

10%
son can interchange a pair in 10 combination out of which only one keps the name unchanged and that is EE
10 combinations for change are
PR
PE
PE
PT
RE
RE
RT
EE
ET
ET

so the probability that despite this interchange, the name remains unchanged = 1/10 = 10%

Is this solution Helpfull? Yes (1) | No   

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