What Is the Correct Way to Calculate the Probability of Flipping Heads?

[mathdad]

Probability of an event happening = (Number of ways the event can happen)/(Total number of outcomes)

Please, explain the above definition

How is the above definition applied to the following question.

A coin is tossed 100 times. How many heads will pop up?

Solution:

Let P = probability

A coin has two sides: heads and tails. The head side is 1/2 of the coin.

P(head pops up) = (2)/100

P(head pops up) = (2) ÷ (100/1)

P(head pops up) = (2) (1/100)

P(head pops up) = 1/50

Correct?

 

[MarkFL]

Probability of an event happening = (Number of ways the event can happen)/(Total number of outcomes)

Please, explain the above definition

How is the above definition applied to the following question.

A coin is tossed 100 times. How many heads will pop up?

Solution:

Let P = probability

A coin has two sides: heads and tails. The head side is 1/2 of the coin.

P(head pops up) = (2)/100

P(head pops up) = (2) ÷ (100/1)

P(head pops up) = (2) (1/100)

P(head pops up) = 1/50

Correct?

If a coin is tossed $0<n$ times, then any number of heads from 0 to $n$ can result. To find the probability of a given number of heads (we'll call this $0\le m\le n$) resulting, we need to use the binomial probability formula:

\(\displaystyle P(X)={n \choose m}\left(\frac{1}{2}\right)^{m}\left(\frac{1}{2}\right)^{n-m}={n \choose m}2^{-n}\)

 

[harpazo]

If a coin is tossed $0<n$ times, then any number of heads from 0 to $n$ can result. To find the probability of a given number of heads (we'll call this $0\le m\le n$) resulting, we need to use the binomial probability formula:

\(\displaystyle P(X)={n \choose m}\left(\frac{1}{2}\right)^{m}\left(\frac{1}{2}\right)^{n-m}={n \choose m}2^{-n}\)

Thank you for helping RTCNTC.

 

[MarkFL]

In the formula, does n = 2 and m = 100?

The other way around...n is the total number of tosses, and m is the number of heads. :D

 

[mathdad]

Is the set up nCm = 100!/2!(100-2)! * 2^(-100)?

 

[MarkFL]

Is the set up nCm = 100!/2!(100-2)! * 2^(-100)?

Yes, there are 100 ways to choose the first flip that will be a heads, and 99 ways to choose the second way there will be a heads, and then we need to divide by the two ways to arrange those two heads, and 2^(100) total outcomes, so we get:

\(\displaystyle P(\text{two heads})=\frac{100\cdot99}{2}2^{-100}=\frac{100!}{2!(100-2)!}2^{-100}={100 \choose 2}2^{-100}\)

 

[mathdad]

I will work this out on paper.

 

[HallsofIvy]

Probability of an event happening = (Number of ways the event can happen)/(Total number of outcomes)

Please, explain the above definition

How is the above definition applied to the following question.

A coin is tossed 100 times. How many heads will pop up?

This is not a probability question and does not have a unique answer.

Solution:

Let P = probability

A coin has two sides: heads and tails. The head side is 1/2 of the coin.

P(head pops up) = (2)/100

P(head pops up) = (2) ÷ (100/1)

P(head pops up) = (2) (1/100)

P(head pops up) = 1/50

Correct?

The correct answer to what question? What does "P(head pops up)" mean? You were told that the probability that a flipped coin will come up heads is 1/2. That is what I would expect P(head pops up) to mean. You can also calculate that the "expected number of heads in 100 flips" is 50. But I don't see where the 1/50 came from and I do not know what question this is supposed to answer.