What is the Probability of Getting Three Heads in 30 Coin Tosses?

[ThunderWolf2016]

Conduct 30 trials and calculate emperical probability of the outcomes. What is the probability of tossing 1 coin three times and getting three heads? Need a flowchart of some sort to show all possible outcomes. This math is new to me and has little understanding of this math lol so any help aanyone can give me will be really appreciative.

 

[MarkFL]

Out of all the possible outcomes from tossing a coin $n$ times, how many of those will be $n$ heads?

For each individual coin toss, how many outcomes are possible? How many of those are heads? Using the fundamental counting principle, how many outcomes will there be for $n$ coin tosses?

 

[MarkFL]

As a follow-up, I will answer my own questions...

Out of all the possible outcomes from tossing a coin $n$ times, how many of those will be $n$ heads?

There is only one outcome in which all tosses will be heads, since there is only 1 way to choose $n$ from $n$:

\(\displaystyle {n \choose n}=\frac{n!}{n!((n-n)!}=1\)

For each individual coin toss, how many outcomes are possible?

There are two possible outcomes, either heads or tails, and for a fair coin, both are equally likely.

How many of those are heads?

Only 1 outcome is heads, therefore we can compute the probability of getting heads on one toss as follows:

We are certain that we will either get heads (H) or tails (T):

\(\displaystyle P(\text{H})+P(\text{T})=1\)

Since we assume the coin is fair, we know:

\(\displaystyle P(\text{H})=P(\text{T})\)

And so we have:

\(\displaystyle P(\text{H})+P(\text{H})=1\)

\(\displaystyle 2P(\text{H})=1\)

\(\displaystyle P(\text{H})=\frac{1}{2}\)

This corresponds with the fact that out of the two possible outcome, only one is favorable:

\(\displaystyle P(\text{H})=\frac{\text{number of ways to get heads}}{\text{total number of outcomes}}=\frac{1}{2}\)

Using the fundamental counting principle, how many outcomes will there be for $n$ coin tosses?

Since for each toss, there are two outcomes, the total number $N$ of outcomes for $n$ tosses is:

\(\displaystyle N=\prod_{k=1}^{n}(2)=2^n\)

And so we may conclude that the probability of getting $n$ heads for $n$ tosses is:

\(\displaystyle P(\text{n heads})=\frac{1}{2^n}\)

So, if $n=3$, the probability of getting all 3 heads is:

\(\displaystyle P(\text{3 heads})=\frac{1}{2^3}=\frac{1}{8}\)

 

[ThunderWolf2016]

thank you all for help is well appreiciated :D