# Discrete Math, counting and probability (white balls and black balls)?

Can anyone help me with this please?

An urn contains ten white balls numbered from 1 to 10, and ten black balls numbered from 1 to 10. A sample of 5 balls is chosen from the urn.

(a) How many di erent samples are there? (b) How many samples in (a) have at least one white ball? (c) How many samples in (a) have the property that the sum of the numbers on the balls is even? (d) How many samples in (a) have the property that the product of the numbers on the balls is even? (e) How many samples in (a) have the property that the numbers on the balls are distinct?

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Sampling without replacement? – Henry Aug 12 '13 at 23:26
Hi Henry, no replacement, please see my answer below and comment... thanks... – Najeeb Aug 12 '13 at 23:37

(a), (b) Your calculations for the first two are right. There are $\binom{20}{5}$ possible samples.

For at least one white, your alternative procedure is more efficient. The answer $\binom{20}{5}-\binom{10}{5}$ is right.

(c) The sum is even if we have an odd number of evens (for then we have an even number of odds). So we want $1$, $3$, or $5$ evens. The number of ways to choose $1$ even and $4$ odd is $\binom{10}{1}\binom{10}{4}$. Write down also the number of ways to get $3$ even and $2$ odd, also $5$ even and $0$ odd, and add.

But we do not need to add! For we get an odd sum if we get $1$ odd and $4$ even, or $3$ odd and $3$ even, or $5$ odd. If you write down the number of ways to do this, you will get exactly the same expression as we got for even sum. So the number with even sum is $\frac{1}{2}\binom{20}{5}$.

(d) There are $\binom{10}{5}$ ways to choose all odd. The rest will give even product.

(e) There are $\binom{10}{5}$ ways to choose $5$ distinct numbers from $10$. For each such choice, look at the numbers you have. Each can be "represented" by a black or a white, giving a total of $\binom{10}{5}2^5$.

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Thanks for the help. – Najeeb Aug 13 '13 at 0:50
You are welcome. The solutions use various themes, which can be reused. – André Nicolas Aug 13 '13 at 0:56

Hints on how to go about each:

a. Consider the possible breakdowns and how many of each can you have? For example, how many ways could you draw 5 white balls, 4 white and 1 black, 3 white and 2 black, ...

b. There are a couple of ways to go here. First, is the consider all the possible ways to select 5 items from a set of 20 and then remove the all black balls cases that would give you the at least one white ball case for a removal way to count this. The flip would be to sum all but the all black ball case in a.

c. This requires a bit more work to consider as you'd want to look at various odd and even cases. While there is the requirement for at least one even, the other 4 balls have to pair up for the sum to be even.

d. This requires something a bit easier in that now you're looking for any even number amongst the 5 balls.

e. This requires you consider which duplicate cases can exist to remove from the overall set in a or else be very careful in constructing a counting case for all the balls to be distinct.

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This is what I got: – Najeeb Aug 12 '13 at 23:26
'There are 20 balls in the urn. 5 balls may be chosen in 20C5 = 20! / 5! 15! = 15,504 b)at least 1 white ball = (10C1)(10C4) + (10C2)(10C3) + (10C3)(10C2)+(10C4)(10C1)+(10C5)(10C0) =2,100 + 5,400+5,400+2,100+252 = 15,252 Or: No white ball = all 5 are black balls. There are (10C0)(10C5) = 252 ways of choosing black balls only. At least 1 white ball = 15,504-252 = 15,252 c) (not sure) half of 15,252 are even = 7,626 d) same as (c) e) 10P5 = 10!/5! = 30,240' – Najeeb Aug 12 '13 at 23:29
Sorry, don't know how to get it look nicer... – Najeeb Aug 12 '13 at 23:29
For a and b, yes I can see that. I'm pretty sure that d isn't the same as c as the case of an odd number of odd numbers as long it isn't all 5 can still produce an even product but not necessarily an even sum. e has the challenge of figuring out how many ways could you get only 3 distinct numbers or 4 distinct numbers as these are the scenarios where you have duplicates. – JB King Aug 12 '13 at 23:34