Probability of having more Heads than Tails tossing a biased coin n times

Question

Suppose the probability of a Head in a single tossing is given by $p\in[0,1]$.

When we toss the coin $n$ times, what is the probability that the number of the total Heads is greater than or equal to the number of the Tails?

My attempt:

$n=1$) One head is required. So, the probability is $p$.

$n=2$) At least one head is required. So, the probability is $p^2+p(1-p)$.

$n=3$) At least two heads are required. So, the probability is $p^3$+${3}\choose{2}$$ p^2(1-p)$.

$n=4$) At least two heads are required. So, the probability is $p^4$+${4}\choose{3}$$ p^3(1-p)$+${4}\choose{2}$$ p^2(1-p)^2$.

$n=5$) At least three heads are required. So, the probability is $p^5$+${5}\choose{4}$$ p^4(1-p)$+${5}\choose{3}$$ p^3(1-p)^2$.

However, I don't know how to simplify and generalize this sequence of sums..

How may I express the probability in a more concise and generalized form?

Answer 1

The compact way to write your sums is $$\sum_{k=0}^\ldots{n\choose k}p^{n-k}(1-p)^k$$ where here $k$ represents the number of tails. So the maximum value of $k$ is when $$k\leq n-k\qquad\text{and}\qquad k+1>n-(k+1)$$ meaning $$k\leq \frac n2\qquad\text{and}\qquad k+1>\frac n2$$ so that $k_{max}=\left\lfloor\frac n2\right\rfloor$:

$$\sum_{k=0}^{\left\lfloor\frac n2\right\rfloor}{n\choose k}p^{n-k}(1-p)^k$$

Answer 2

If $X$ is the number of heads then $X$ has Binomial distribution with parameters $n$ and $p$. For $n$ even the required probability is $P(X >\frac n 2)$ and for $n$ odd it is $P(X\geq \frac {1+n} 2)$. Use the fact that $P(X=k)=\binom {n} {k} p^{k}(1-p)^{n-k}$.