Probability Question.

Maybe you want to calculate what the minimum and maximum sum can be from Tina's set. (nudge)

I took the probability that Steve has to pick a number (.1 for set{1,2,3...10}) times the probability that Tina would have a smaller number than that and added it up for each number within the set.

Ie.
= (.1*1) + (.1*.9) + (.1*.8)+(.1*.6)....

Is this correct?

P(Steve having a number larger than Tina's sum) = .4

I'm afraid I'm a long time out of school, so you might not want to take my advice. But I would come up with a set of possible sums (I get 3-9 inclusive) and add up the probability that Steve will pick a number greater than each sum times the probability that Tina will choose numbers that add up to each sum.

No... this is not the correct direction to go in.

The key (and this is a big nudge) is the word "sum". You first need to find all of the possible "sums" that Tina could get (there aren't that many)... and the probability for each one.

There is 125/250 chance that Steven’s number is larger or equal to the sum of Tina's numbers.
There are 1+2+3+4+5+4+3+2+1+0=25 possibilities in which the numbers are equal.
125/250-25/250=100/250=?

I don't know how you are supposed to do this formally, and I'm sure there's a shorter way to do it, but that would mean having to learn it instead of just figuring it out. So here's how I'd do it with the caveat that if this is school work I'm sure they are looking for something very different.



Possible combinations:

1+2 = 3
1+3 = 4
1+4 = 5
1+5 = 6
2 +3 = 5
2+4 = 6
2+5 = 7
3+3 = 6
3+4 = 7
3+5 = 8
4+5 = 9

So of those 11 combinations the sums have the following odds:

3 = 1/11
4 = 1/11
5 = 2/11
6 = 3/11
7 = 2/11
8 = 1/11
9 = 1/11



10 possibilities.

So there's a 1/11 chance that Tina's sum is 9 and a 1/10 chance that Steven's number is 10.

1/11 x 9/10 = 9/110 so far

There's a 1/11 chance her sum is 8 and a 2/10 chance that Steven's number is greater than 8 so:

1/11 x 8/10 = 8/110 or 17/110 so far.

There is a 2/11 chance her sum is 7, and a 3/10 chance Steven's number is greater so:

2/11 x 7/10 = 14/110 or 31/110 so far

There is a 3/11 chance her sum is 6 and a 4/10 chance Steven's number is greater so:

3/11 x 6/10 = 18/110 or 49/110 so far.

There is a 2/11 chance her sum is 5 and a 5/10 chance Steven's number is greater so:

2/11 x 5/10 = 10/110 or 59/110 so far.

There is a 1/11 chance her sum is 4 and a 6/10 chance Steven's number is greater so:

1/11 x 4/10 = 4/110 or 63/110 so far.

There is a 1/11 chance her sum is 3 and a 7/10 chance Steven's nubmer is greater so:

1/11 x 3/11 = 3/110 or 66/110 (or 40% that his is greater).

That's the road I was going down too, except that I got 10 possible outcomes from Tina's selection. I think I was assuming that she couldn't pick the same number twice.

I got:

(1/10)(7/10) + (1/10)(6/10) + (2/10)(5/10) + (2/10)(4/10) + (2/10)(3/10) + (1/10)(2/10) + (1/10)(1/10)

for a total of .4. So a 40% chance that Steve will pick a number greater than the sum of Tina's numbers. So I guess it comes out the same. This is also the number that babemomlover got, so it looks like there are a lot of ways to get the correct answer.

IMO the simplest method is to sketch out a 2D sample space with the numbers 1 - 10 on the x axis and the eleven sum scores 3 - 9 on the y axis. Tick and count the squares where x>y giving a fraction 44/ 110.
p=0.4

you might use a negative on steve to tina

From combinations you know Tina has 5!/2!/3!=10 possibilities
possibilities 1 of sum 3--1,4--2,5--2,6--2,7--1,8--1,9

Tina has 1/10 of 3, 1/10 of 4, 2/10 of 5, 2/10 of 6, 2/10 of 7, 1/10 0f 8, & 1/10 of 9

Steve has the possibility of 3/10 chance of not beating a Tina 3, and so fourth. now putting them all together and betting that Steve does not beat Tina the

P(E)=1/10*3/10+1/10*4/10+2/10*5/10+2/10*6/10+2/10*7/10+1/10*8/10+1/10*9/10
or
P(E)=(3+4+10+12+14+8+9)/100=60/100=.6

Since the outcome is a win or a loss (no ties) and Steve has a point six chance of losing then he has a 0.4 chance of winning.

Rap

I'd use a more general form. The average of all the numbers in data set one is 3, so the average of two numbers would be 6. The probability of picking a number greater than six from the second set is 40% (7-10). This works really well if the numbers are evenly distributed. The long form has to be used if the number spacing is not uniform.

"This works really well if the numbers are evenly distributed. The long form has to be used if the number spacing is not uniform."

Which makes that more special than general...