If a fair coin is flipped 5 times, what is the probability of getting exactly 3 heads

Question

Q: "If a fair coin is flipped 5 times, what is the probability of getting exactly 3 heads?"

A: we can do this question by drawing blanks strategy. for eg. $$\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)*10$$

but is there a way to do this same problem using combinations? for eg. $$\dfrac{_5C_3}{_{10}C_5} = \dfrac{10}{252}$$ {so how should I get $\dfrac{10}{32}$ instead of $\dfrac{10}{252}$?}

I am really not sure how to do this using combinations. Any help would be greatly appreciated.

Thanks

EDIT: sorry, I edited the question from my previous question modified question. This is the original question. my bad.

Answer 1

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.

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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.

Answer 2

Taking red balls as "success", it is a binomial distribution.

Put $p=\frac{1}{2}$ (red) and $q=1-p$ (blue), $n=5$ and $k=3$ in

$$ \binom{n}{k}p^kq^{n-k}. $$

This reads "the probability of $k$ successes after $n$ independent trials". Since you replace the balls, each trial is an independent events.

Answer 3

Here is how you do it using combinations.

There are $\binom53$ sequences of five coin tosses with exactly three heads. The probability of each of these is $(1/2)^5$. Therefore, the probability of three heads is $\binom53(1/2)^5.$

Your method using $\binom{10}5$ is incorrect. Nowhere are there five objects being chosen from ten.

Answer 4

The logic you have applied to answer this question is faulty. There are totally 5 trials, and each trial will give either Head or Tail. And what you have done is that you have made a sample space by combining all the elements, you have put all the possible 5 heads and 5 tails. You can't simply create a sample space by putting all the elements of 5 trials together. Rather, the actual sample space should be like {HHHHH, TTTTT, HHHHT,.... ,},
not, {H, H, H, H, H, T, T, T, T, T} which you have done. Now, it can be solved by combination.