Probability
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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

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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
 ()

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