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(#M40012730) TCS question Keep an EYE

find probablity of a*b+c is even . numbers are from (1,2,3,4,5). digits can be repeated.

Asked In TCS SHAIK NAYAB RASOOL (11 years ago)

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venkatesh bollam  (11 years ago)

given set is 1,2,3,4,5
our aim is to find probability so that a*b+c should be even
a*b+c gives even number when two odd numbers is added
i.e, odd no+odd no gives even no.
and even no + even no gives even no
so lets consider case 1: odd+odd=even (a*b) is odd and c is odd
now a*b gives a odd no when a and b are odd no.
probability of odd number from given set=3 by 5
so case 1 gives the prob as (3 by 5)*(3 by 5)*(3 by 5)=(27 by 125)........1

case 2: even+even=even
prob of even no from the set=2 by 5
a*b is even and c is even (three cases are possible)
a(odd) b(even) c(even)= (3 by 5)*(2 by 5)*(2 by 5)and
a(even) b(odd) c(even)= (2 by 5)*(3 by 5)*(2 by 5)and
a(even) b(even) c(even) = (2 by 5)*(2 by 5)*(2 by 5).
so totally (27+12+12+8)by 125......ans

Is this solution Helpfull? Yes (35) | No (1)  

snehil  (11 years ago)

for make a even number there will be the case
1) a even b even c even ie prob = (2/5)*(2/5)*(2/5)=8/125
2) a even b odd c even ie prob = (2/5)*(3/5)*(2/5)=12/125
3) a odd b even c even ie prob = (3/5)*(2/5)*(2/5)=12/125
4) a odd b odd c odd ie prob = (3/5)*(3/5)*(3/5)=27/125
so tltal probability =(8+12+12+27)/125 ans

Is this solution Helpfull? Yes (19) | No (1)  

preeti arya  (11 years ago)

its 59/125.... correct

Is this solution Helpfull? Yes (5) | No (3)  

pihu  (11 years ago)

ans should be 99/125
total cases =125
cases for
1- a (even ),b+c(even)
a=2
b+c=(3c1*3c1)+(2c1*2c1)=13
total cases=2*13 =26

2- a (odd ),b+c(even)
a=3
b+c=(3c1*3c1)+(2c1*2c1)=13
total cases=3*13 =39

3-a (even ),b+c(odd)
a=2
b+c=(3c1*2c1)+(3c1*2c1)=12
total cases=2*12 =24

now, p= (24+39+26)/125 = 99/125

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RAVINDRA KUMAR YADAV  (11 years ago)

probability of even number=2/5
prob of odd number=3/5
total prob=(2/5)*(3/5)*(2/5)+(3/5)*(2/5)*(2/5)+(3/5)*(3/5)*(3/5)=51/125

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