Latest TCS Aptitude Question SOLUTION: chosen a,b,c randomly and with replacement from the set {2,3,4,5,6}the probablity that a*b+c is even ?

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(#M40019199) TCS question Probability Keep an EYE

chosen a,b,c randomly and with replacement from the set {2,3,4,5,6}the probablity that a*b+c is even ?

Asked In TCS Shiva (10 years ago)

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gangadhar  (10 years ago)

ANSWER=18/25

total number=5 numbers are {2,3,4,5,6}
probability to pick a number p=1/5
even numbers=3 {2,4,6} and Odd numbers=2 {3,5}
probability of even number P.E=3/5
probability of even number P.O=2/5
given A*B+C is even
case 1:
if A is even then B can be either even or odd and C must be even to give even number
P.E*p+P.O=
3/5*1/5+3/5=18/25
therefore probability=18/25
case 2:
if A and B are Odd then C must be odd to give an even number
this case fails because there are only 2 odd numbers
probability=0

the probablity that a*b+c is even is p(case 1)+ p(case 2)=18/25+0= 18/25
probability =18/25

Is this solution Helpfull? Yes (31) | No (4)  

tanvi agarwal  (10 years ago)

There are 4 cases possible
P(e)=3/5 and P(o)=2/5

Case1-All a,b,c are even
so P1= (3/5)^3
Case2- a,c = even and b=odd
so P2=(3/5)(2/5)(3/5)
Case3- b,c=even and a=odd
so P3=(2/5)(3/5)(3/5)
Case4- all odd
so P4=(2/5)^3

Adding all to get the total probability =71/125

Is this solution Helpfull? Yes (8) | No (6)  

Shiva  (10 years ago)

See number to be even...>
{even*even+even}
{even*odd+even}
{odd*even+even}

here, P(e)=3/5,P(o)=2/5
so total probablity is.....>
(3/5*3/5*3/5)+(3/5*2/5*3/5)+(2/5*3/5*3/5) = 63/125............hope it's correct.

Is this solution Helpfull? Yes (7) | No (3)  

Pankaj Bawane  (10 years ago)

Probability of chosing a,b,c randomly and with replacement = 1/5
Thus, a*b+c = 1/5*1/5+1/5 = 6/25

Is this solution Helpfull? Yes (1) | No (7)  

sivashankari  (10 years ago)

(a,b,c)= (2,4,6)
2*4+6=14
2*6+4=16
4*6+2=26 all are even

Is this solution Helpfull? Yes | No (3)  

shivangi jain  (10 years ago)

Total ways of selecting 3 no.=5c3=10
The sum willbe odd in 2 cases
1_when all 3 nos r even whose probablity is 1
2-when2 no. r odd nd 1 is even since only 2 no. r odd in this to select 1 even from 3 even no.s
i.e. 3c1=3
fAvourable cases =3
thus p=4/10 ans=2/5.

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ABHISHEK KUMAR MISHRA  (10 years ago)

no of event=3C2*2C1 + 3C3
no of sample space=5C3
ans=7/10

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suganya  (10 years ago)

Answer=7/27
there are 3 cases
p1=all are even p2=a and c are even p3=b and c are even
case 1:all are even
p1=(1/3)*(1/3)*(1/3)=(1/27)
p2=(1/3)*(1/3)=(1/9)
p3=(1/3)*(1/3)=(1/9)
p=p1+p2+p3=7/27

Is this solution Helpfull? Yes | No (1)  

karthik  (10 years ago)

1-(6/pow(5,3))
3 and 5 are odd. So, odd*odd+even=odd which is (a,b) either {(3,5),(5,3)}

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sami wardhan karn  (10 years ago)

((1/2)*(1/2))+1/3=7/12

Is this solution Helpfull? Yes | No (2)  

urmi bose  (8 years ago)

71/125 is the correct answer
As there are 4 cases:
1.all odd ie (2/5)^ 3
2.all even ie (3/5)^3
3.a= odd other two even ie (3/5)^2* 2/5
4.b= odd other two even ie (2/5)^2* 3/5
Adding all we get 71/125

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