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(#M40028002) TCS Verbal Ability question tcs Keep an EYE

a, b, c are chosen randomly and with replacement from the set {1,2,3,4,5}. Find the probability that a*b+c is even.

Asked In TCS rakesh (11 years ago)

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Read Solution (3) :

Kundan Kumar  (11 years ago)

Firstly:-
Proba.. of getting even number = 2/5
Proba. of getting odd number= 3/5
Now Considering
ei(i=1,2,3)= number when it is taken as even
Oi(i=1,2,3)= number when it is taken as odd
a ( e1,o1)
b ( e2,o2)
c (e3,o3)
Therefore,
Eight possible cases(just find the cross product of a,b,c)
{ e1e2e3, e1o2e3, o1e2e3, o1o2e3,e1e2o3,e1o2o3,o1e2o3,o1o2o3)
Among them 4 possible cases results to an even number
ie.
even* even + even ==> even
even*odd+ even ==> even
odd* even + even==> even
odd * odd + odd ==> even
Since replacement is allowed :

(2/5)*(2/5)*(2/5) + (2/5)*(3/5)*(2/5) + (3/5)*(2/5)*(2/5) + (3/5)*(3/5)*(3/5)
= 59/125 Ans.

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ddev  (11 years ago)

@kundan: (2/5)*(2/5)*(2/5) is not possible because in question there are only two even numbers.. (2/5)*(3/5)*(2/5) + (3/5)*(2/5)*(2/5) + (3/5)*(3/5)*(3/5)= 51/125 would be correct..

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MIDHUN K  (5 years ago)

replacement is ok so 4 possible combination is possible

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