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.
Best Answer
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.