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

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The probability of getting three heads in three coin tosses is calculated using the formula P(n heads) = 1/(2^n), resulting in P(3 heads) = 1/8. Each coin toss has two possible outcomes: heads or tails, making the total outcomes for n tosses equal to 2^n. The discussion emphasizes that there is only one way to achieve n heads in n tosses, confirming the calculation. The fundamental counting principle is applied to derive these probabilities. Understanding these concepts is essential for grasping basic probability in coin toss scenarios.
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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.
 
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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?
 
As a follow-up, I will answer my own questions...

MarkFL said:
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 $n$:

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

MarkFL said:
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.

MarkFL said:
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):

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

Since we assume the coin is fair, we know:

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

And so we have:

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

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

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

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

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

MarkFL said:
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:

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

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

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

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

$$P(\text{3 heads})=\frac{1}{2^3}=\frac{1}{8}$$
 
thank you all for help is well appreiciated :D
 
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