Probability binomial the A fair coin is thrown ten times. Find the probability that more heads arise than tails?
Hi guys my question is a binomial theorem one I know that I'm just finding it hard to know what goes where. I get that in order to have more heads I must have 6 heads or more nut find it there to use the. Formula, thanks
 A: Hint: Actually there are three possible outcomes, when tossing a coin $10$ times, which are the following (and with their respective probabilities): 


*

*More heads than tails: with probability $p$ (this is the probability you are looking for).

*More tails than heads: with the same probability, i.e. $p$ (due to symmetry!).

*Equal number of tails and heads: That is the probability of $5$ heads, i.e. $$\dbinom{10}{5}\left(\frac12\right)^5\left(1-\frac12\right)^{10-5}$$


Now, these three probabilities add up to $1$ ...
A: This is a binomial setting as you've mentioned. What we're basically looking to find is $P(H>5)$, or the probability of getting more than $5$ heads. This is what "More heads than tails" means. As you know, when we do bionmcdf(n,p,k), or use the equation, it's more convenient to change this to a probability with $0$ as our lower bound. Let's turn that $P(H>5)$ into:
$$P(H>5)=1-P(H\le5)$$
Cool. Let's look at the different variables we'll need for a binomial distribution. $n=\text{number of trials, } p=\text{probability of "success", } k=\text{number of "successes"}$. Guess what? We know everything! $n=10$, because we're tossing $10$ coins. $k=5$, because we're looking for $P(H\le5)$ so we want MAX $5$ successes, and $p=\frac{1}{2}$ because there's a $50\%$ chance of getting a head.
We can just plug it into the calculator, and we'll get: 1-binomcdf(10,0.5,5)=37.6953%. Equation-wise, recall that this is our equation:
$$P(X\le k)=\sum_{k=0}^{k} {n\choose k}p^k(1-p)^{n-k}$$
I am used to using the variable $k$ twice, a little confusing but will do. For this problem, it will be:
$$P(H>5)=1-P(H\le5)=1-\sum_{k=0}^{5} {10\choose k}\left(\frac{1}{2}\right)^k\left(\frac{1}{2}\right)^{10-k}$$
This will come out to be $\approx 37.6953\%$ as well.
I did this one-sided because bioncdf(n,p,k) is one-sided, but you can do it like this too:
$$P(H>5)=\sum_{k=6}^{10} {10\choose k}\left(\frac{1}{2}\right)^k\left(\frac{1}{2}\right)^{10-k}$$
I started from $6$ because it doesn't include $5$ (greater than). You get the same answer.
Cheers!
-Shahar
