# Easy Probability Solved QuestionAptitude Discussion

 Q. If $x$ is chosen at random from the set {1,2,3,4} and $y$ is to be chosen at random from the set {5,6,7}, what is the probability that $xy$ will be even?
 ✖ A. 5/6 ✖ B. 1/6 ✖ C. 1/2 ✔ D. 2/3

Solution:
Option(D) is correct

$S$ ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)}

Total element $n(S)=12$

$xy$ will be even when even $x$ or $y$ or both will be even.

Events of $x, y$ being even is $E$.

$E$ ={(1,6),(2,5),(2,6),(2,7),(3,6),(4,5),(4,6),(4,7)}

$n(E) = 8$

So, Probability,

$P=\dfrac{n(E)}{n(S)}$

$P=\dfrac{8}{12}$

$P=\dfrac{2}{3}$

## (7) Comment(s)

Pam
()

Hey guys please help me out. Tell me when to consider xy as xXy(2x5=10)multiples and when xy as xy(25)number in questions like these. because if its not mentioned precisely then its a bit difficult to get the right answer as the options satisfies both the above conditions.

Ram Kr Prajapati
()

Previously I was wrong because I thought $xy$ as two digits number whose unit place is $y$ and tens place is $x$.

Sorry correct answer is $\dfrac{8}{12}=\dfrac{2}{3}$

Ram Kr Prajapati
()

I think $xy$ is even if $y$ is even. I think in your solution how you consider these numbers (2,5), (2,7),(4,5),(4,7).

Ziah
()

For, $(x,y)=(2,5)$, we get $xy=2\times 5 = 10$, which is an even number. And the same is true for other numbers as well.

So even if either of $x$ or $y$ is even or both of them are even, it satisfies the condition.

Hope that helps.

Ram Kr Prajapati
()

I think the answer should be $\dfrac{4}{12}= \dfrac{1}{3}$

Ziah
()

Why do you feel so?

I believe the solution given is correct. If you explain more than probably I could understand your concern better.

Jbhj
()

wicked wasome