# Finding Number of cases

Suppose we have 20 cards numbered 1 through 20 and we want to pick 5 of them at the same time

a.How large is our total sample space ?

b.What is the probability of numbers 1 and 2 are not in our sample?

c.What is the probability of at least one card number is 3 , 4 ?

for a I think C(20,5)

I have no idea for b and c

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A natural sample space is the collection of all "hands" of $5$ numbers. There are $\dbinom{20}{5}$ such hands, and they are all equally likely.

For b), there are $\dbinom{18}{5}$ ways to choose $5$ cards, none of which is $1$ or $2$. So our probability is $$\frac{\binom{18}{5}}{\binom{20}{5}}.$$ If desired, considerable simplification is possible.

For c), the simplest way is to subtract the answer to b) from $1$. This is because the event "at least one from $1$ or $2$" happens precisely if the event described in b) doesn't happen.

We can also solve the problem in a way close to the idea you mentioned in a comment. If we want a $1$ or a $2$ or both, there are three ways this can happen: (i) $1$ but not $2$; (ii) $2$ but not $1$; both $1$ and $2$.

The number of choices for (i) is $\dbinom{18}{4}$. The number of choices for (ii) is also $\dbinom{18}{2}$. And the number of choices for (iii) is $\dbinom{18}{3}$. Add up, and for the probability divide by $\dbinom{20}{5}$.

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No, please see what I added for c) – André Nicolas Nov 18 '12 at 3:52

Your answer of $\binom{20}5$ for (a) is correct.

(b) To get the probability that $1$ and $2$ are not in the sample, you must count the $5$-card sets that don’t contain either $1$ or $2$. These are the $5$-card sets chosen from $\{3,4,5,\dots,20\}$; how many of them are there? This is really a problem just like (a). Once you have that, what fraction must you form to get the desired probability?

(c) Now observe that the events in (b) and (c) are complementary, so their probabilities add up to a known amount; what is that amount? Use it and your answer to (b) to get the answer to (c).

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