Unfair coin question

Question

Supposed that you flip an unfair coin with the probability of heads being $p({\rm heads}) = 3/4$ and the probability of tails being $1/4$ a total of ten times. How do you find the possibility of getting exactly $x$ heads and at least $x$ heads?

Answer 1

In general, if the probability of heads on a given toss is $p$, and you toss $n$ times, the probability of getting exactly $x$ heads is $$\bbox[10px,#ffd]{P(\text{exactly } x\text{ heads})=\binom{n}{x}p^x (1-p)^{n-x}},\quad x=0,1,\ldots,n.$$

To get the probability of getting at least $x$ heads, you sum up the probability of getting exactly $x$ heads, $x+1$ heads, ..., $n$ heads. So

$$\bbox[10px,#ffd]{P(\text{at least }x\text{ heads}) =\sum_{k=x}^{n}\binom{n}{k}p^k(1-p)^{n-k}}.$$

If you want probabilities for tails instead of heads, replace every $p$ above with $1-p$. Note that $n=10$ and $p=3/4$ in your particular example.

By the way, the number of heads you get is said to follow a binomial distribution. See the Wikipedia page for more information.

Answer 2

Hint: use the binomial distribution. See, in particular, the examples on the Wikipedia page.

Answer 3

The probability of getting exactly $x$ tails is $\binom {10} {x} \left (\frac 3 4 \right )^x \left ( \frac 1 4 \right )^{10-x}.$

The probability of getting at least $x$ many heads is $\sum\limits_{k=x}^{10} \binom {10} {k} \left (\frac 3 4 \right )^k \left ( \frac 1 4 \right )^{10-k}.$