Probability of getting exactly 2 heads in 3 coins tossed with order not important?

Question

I have been thinking of this problem for the post 3-4 hours, I have come up with this problem it is not a home work exercise

Let's say I have 3 coins and I toss them, Here order is not important

so possible sample space should be

0 H, 1 H, 2 HH, 3 HHH (H being heads) TTT, HTT, HHT, HHH

since P(T) and P(H) =1/2;

Here we have fair coins only, Since each and every outcome is equally likely, answer should be

1/4 (is this correct)

and if that is correct, all of the probabilities don't add up to one, will I have to do the manipulation to make it add up to one, or I am doing anything wrong.

EDIT In my opinion, with order being not important, there should be only 4 possible outcomes. All of the answers have ignored that condition.

Answer 1

Consider all the possible ways to get two heads, $\rm HHT, HTH \; and \; THH$. There are $2 \cdot 2 \cdot 2 = 8$ possible combinations in total. Therefore, the answer is $3/8$.

Your answer is wrong because the number of ways of changing around $\rm HHT$ (3) is not the same as the number of ways of changing around $\rm HHH$ (1). Can you see why this would invalidate your argument?

General solution: Binomial distribution. The probability of getting $k$ successes (here $2$) in $n$ trials (here $3$) is given by:

$$ \Pr(x=k) = \binom{n}{k} p^k (1-p)^{n-k}$$

Where $p$ is the probability of success (here, $p=1/2$), and $\binom{n}{k} = n!/(k!(n-k)!)$. This gives us:

$$ \binom{3}{2} \left(\frac12\right)^2 \left(1 - \frac12\right)^{3-2} $$ $$ 3 \cdot \frac14 \cdot \frac12 $$ $$ \frac38 $$

Answer 2

The sample space has size $2^3 = 8$ and consists of triples $$ \begin{array}{*{3}{c}} H&H&H \\ H&H&T \\ H&T&H \\ H&T&T \\ T&H&H \\ T&H&T \\ T&T&H \\ T&T&T \end{array} $$

The events $$ \begin{align} \{ 0 \text{ heads} \} &= \{TTT\}, \\ \{ 1 \text{ head} \} &= \{HTT, THT, TTH\}, \end{align} $$ and I'll let you figure out the other two.

The probabilities are, for example, $$ P(\{ 1 \text{ head} \}) = \frac{3}{8}. $$

This is called a binomial distribution, and the sizes of the events "got $k$ heads out of $n$ coin flips" are called binomial coefficients.

Answer 3

The outcomes you are looking for are either THH, HTH or HHT. Taking a look at for example THH: the possibility to toss T or H is $0.5$. Thus the possibility to throw T and then H and then H is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$. But since we have three ways to "achieve" the desired result, the possibility of throwing exactly tow heads in three tosses is $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$.

If you continue you like this, you'll find that the possibilites of all possible outcomes (THH, TTH, TTT, HTT ...) add up to 1 indeed.

Answer 4

Saying "order is not important" does not mean that each scenario (0, 1, 2, 3 heads) is equally likely. The probability of getting one head, for example, means that you got one head first, one head second, or one head third. There are 8 possible outcomes, so the probability of one head must be 3/8.