Question

Algebra/Probability
Two distinct integers, x and y, are randomly chosen from the set {1, 2, 3, 4, 5, 6, 7, 8, 9 ,10}. What is the probability that xy-x-y is even?

Answer 1

I think splitting this up into some cases will help.

Case 1: Suppose x and y are even. Then \(xy-x-y\) is also even, since (even\(\times\)even=even) and subtracting evens from an even gives an even.

Case 2: Suppose x and y are odd. Then \(xy-x-y\) is even, since (odd\(\times\)odd=even) and subtracting an odd twice gives an even.

Case 3: Suppose x is even and y is odd. Then \(xy-x-y\) is even, since (odd\(\times\)even=odd). Subtracting an odd gives an even, then subtracting an even gives another even.

Looks like \(xy-x-y\) is even no matter what you pick for \(x\) and \(y\).

Answer 2

But the probability is not 1

Answer 3

I tried it, and it told me it was wrong.

Answer 4

Maybe the accepted answer is 100%?

Answer 5

xy-x-y+1 should be odd
(x-1)(y-1) should be odd

Answer 6

I don't understand how that works into the problem

Answer 7

both x and y must be odd, right ?

Answer 8

@hartnn is saying that the difference between an even and an odd number (consecutive numbers) is 1.

Add one to \(xy-x-y\) and you have \(xy-x-y+1\), which is odd. Which means \(xy-x-y\) is even. I'm not misinterpreting am I?

Answer 9

thats correct. you are not misintrepreting

Answer 10

Ah, I see

Answer 11

so only one case, both x and y must be odd, the probability of which is ?

Answer 12

@wgary, hartnn's right, I made a mistake with my second case. (odd\(\times\)odd=odd\), not even. Subtracting two odds gives another odd. Sorry

Answer 13

Hmm, so does that mean the probability would be 2/3?

Answer 14

Wait, no, because there are uneven probabilities of each case occurring, right?

Answer 15

I think the probability would be\[\frac{\text{(number of pairs of odds)}}{\text{(number of possible pairs)}}\]

Answer 16

how u got 2/3 ?

Answer 17

Or equivalently \(\dfrac{5\times4}{10\times9}\)

Answer 18

i think it this way :
choosing one odd from 10 -----> 5/10 = 1/2
AND choosing another odd from remaining nine,
4/9
total = 1/2* 4/9 = .... ?

Answer 19

And so I'm assuming that means the probability would be 2/9? (of getting an even number)

Answer 20

There are 4 cases with equal probabilities, three of which are even.
even-even = even
odd-odd = odd
even-odd = even
odd-even = even
can you now figure out the probability of even?

Answer 21

well, i'd say don't assume, prove it like we did :)

Answer 22

@SithsAndGiggles already did most of the work with the four cases plus your correction.
All we have to prove is that the probability of getting P(even) from {1,2,3,4,5,6,7,8,9,10} equals 5/10.

Answer 23

Just to make sure, (with SithsandGiggles method), it would be 5 odds, then 4 odds for the next choice, and then 10 (even or odd number) and 9 for (even or odd number)

Answer 24

@wgary
We are choosing from the same pool of 10 numbers for both x and y. So the probability of choosing an even is the same for each of x and y.

Answer 25

They're not being replaced though. x and y are distinct.

Answer 26

I thought they were distinct numbers, so they couldn't be the same

Answer 27

yup, 'distinct' is the keyword, they can't be same

Answer 28

Oh, sorry, didn't read properly the question, my bad!

Answer 29

Also, I think another way to think of it would be 5c2/10c2

Answer 30

In that case, there will be 8 cases, the four sithsAndGiggles cases multiplied by odd first and even first. Symmetry tells me that the results are not changed.

Answer 31

C (5,2) = 10
C (10,2) = 45
yes,
5C2 /10C2 will work :)

Answer 32

I guess I goofed again. The probability of odd-odd is (5/10)*(4/9)=20/90=2/9
So P(even)=1-P(odd)=7/9.

Answer 33

@SithsAndGiggles As hartnn said, 5c2 = 10, and 10c2 =45

Answer 34

P = favorable events/ total events
= choosing 2 odds from 5 / choosing 2 numbers from 10 = 5c2/10c2
:)

Answer 35

Right, I was thinking of C(5,1) and C(10,1)... Sorry about that

Answer 36

Well, thanks to everyone ;)

Answer 37

welcome ^_^