We're asked for the probability of getting EXACTLY three heads on five flips of a fair coin. This question can be approached in a couple of different ways, but they all involve a bit of 'Probability math.'
To start, since each coin has two possible outcomes, there are (2)(2)(2)(2)(2) = 32 possible outcomes from flipping 5 coins. To find the number of outcomes that are EXACTLY 3 heads, you can either use the Combination Formula or do some 'brute force' math and map out all of the possibilities.
By choosing 3 heads from 5 tosses, we can use the Combination Formula: N!/(K!)(N-K)! = 5!/(3!)(5-3)! = (5)(4)(3)(2)(1)/(3)(2)(1)(2)(1) = (5)(4)/(2)(1) = 10 possible ways to flip 3 heads from 5 tosses.
You could also list out the options:
HHHTT
HHTHT
HTHHT
THHHT
HHTTH
HTHTH
THHTH
HTTHH
THTHH
TTHHH
Either way, you have 10 total options that fit what we're looking for out of a total of 32 outcomes. 10/32 = 5/16
First, a key element of the problem is that you have a set of possible outcomes from the sequence of coin flips, and each is equally likely. This is true only because you have a fair coin, that is, it does not favor heads and does not favor tails.
It's because of the fact that you have a fair coin that you can even think about using a method where you put the number of "desired" outcomes in the numerator of a fraction and put the total number of possible outcomes in the denominator. But since you have a fair coin, you can use that method.
Now let's count the number of outcomes to put in the denominator.
A little notation will help here. Let $H_n$ represent the fact that the coin came up heads on the $n$th flip. Let $T_n$ represent the fact that the coin came up tails on the $n$th flip.
You have noticed $10$ different things that might occur during the sequence of flips: $$ H_1, T_1, H_2, T_2, H_3, T_3, H_4, T_4, H_5, T_5. $$ That is, there are five flips, you could get a head on any of those five flips, and you could get a tail on any of those five flips.
It is also true that among these $10$ things that could happen, in the end there will be exactly $5$ of them that actually did happen. You might think that this means there are $_{10}C_5$ possible outcomes for the sequence of flips--that is, choose $5$ of the $10$ things in the list.
But $_{10}C_5$ counts the number of ways to choose $5$ items from $10$ if it is possible to choose any five things from the list. Given the list of $10$ things, $ H_1, T_1, H_2, T_2, H_3, T_3, H_4, T_4, H_5, T_5, $ here is one set of $5$ things we can choose from that list:
$$ H_1, T_1, H_2, T_2, H_3. $$
What does this signify? It says the first flip came up heads but the first flip also came up tails. Likewise for the second flip. The fourth and fifth flips did not produce any results at all. Coins do not work that way.
The idea of "choose $5$ out of $10$" therefore has you counting impossible outcomes as possible outcomes. That's wrong. So $_{10}C_5$ is simply not a correct way to count the possible number of outcomes of five coin tosses.
The actual number of outcomes is $2^5 = 32.$ This is a small enough number that you can actually write every possible sequence of coin tosses on a single piece of paper in not too much time. It seems this would be a useful exercise at this point in your studies. Make sure you write each possible sequence of tosses (e.g. $H_1T_2H_3H_4H_5$) exactly once, and count them. It helps if you write the list in a sequence with a pattern that helps you make sure you wrote every sequence and did not write any twice.
If you count the number of outcomes that way, you get the result $$ \frac{_5C_3}{2^5} = \frac{10}{32},$$ which is the same thing you got with the "drawing blanks" method. Note that this way of figuring the probability does use combinations in the one place where it really makes sense to use combinations.
By the way, the answer comes out exactly the same whether you flip a fair coin $5$ times or draw $5$ times with replacement from a bag with an equal number of red and blue balls. The number of balls in the bag has no impact on the result as long as exactly half are red and half are blue. You are equally likely to get red or blue on the first draw, and in either of those cases, since you put the ball back in the bag before drawing again (that's what "with replacement" means), you again have equal numbers of red and blue balls before the second draw and you are equally likely to get red or blue.
The number of possible outcomes of each coin flip is 2 (either heads or tails.) So the probability of either a heads or a tails is 1/2.
In Chapter 2 you learned that the number of possible outcomes of several independent events is the product of the number of possible outcomes of each event individually. So the number of combinations that 2 coin flips will give you is: 2 x 2 = 4. In this case we are flipping 5 coins -- so the number of possibilities is: 2 x 2 x 2 x 2 x 2 = 32.
In case you'd like to scrutinize this point, here is the list of all possible combinations:
Coin 1 2 3 4 5 outcome 1 h h h h h (5 h) 2 h h h h t (4 h) 3 h h h t h 4 h h t h h 5 h t h h h 6 t h h h h 7 h h h t t (3 h) 8 h h t h t 9 h t h h t 10 t h h h t 11 h h t t h 12 h t h t h 13 t h h t h 14 h t t h h 15 t h t h h 16 t t h h h 17 t t t h h (2 h) 18 t t h t h 19 t h t t h 20 h t t t h 21 t t h h t 22 t h t h t 23 h t t h t 24 t h h t t 25 h t h t t 26 h h t t t 27 t t t t h (1 h) 28 t t t h t 29 t t h t t 30 t h t t t 31 h t t t t 32 t t t t t (0 h) (Whew!)As you can count for yourself, there are 10 possible ways to get 3 heads. Thus, the probability of getting 3 heads from 5 coin flips is: 10/32, or 5/16.