# Probability mass function for a coin toss

• XodoX
In summary: If not, then please explain how you arrive at that 0.5 probability.In summary, the probability of getting heads on a single toss is 0.5.
XodoX

## Homework Statement

A biased coin is tossed ten times. Suppose that the probability of getting heads on a single toss is p. Let X be the number of heads obtained.
a)Give an algebraic formula for the probability mass function of X.
b) What do you think E[X] should be.

## The Attempt at a Solution

No idea. The probability mass function for a discrete random variable is:

pX(x)=Pr(X=x) for all x

I suppose that's what they want for a), but one for this example ? No idea about b), either. It's a biased coin toss. How am I supposed to find this out here? Otherwise, it would be E[X]=0.5

Since they don't tell what you what the bias is, you can kind of ignore it. All you know is that the probability is p. What if you tossed a fair coin 10 times? What is the probability of getting x heads?

XodoX said:

## Homework Statement

A biased coin is tossed ten times. Suppose that the probability of getting heads on a single toss is p. Let X be the number of heads obtained.
a)Give an algebraic formula for the probability mass function of X.
b) What do you think E[X] should be.

## The Attempt at a Solution

No idea. The probability mass function for a discrete random variable is:

pX(x)=Pr(X=x) for all x

I suppose that's what they want for a), but one for this example ? No idea about b), either. It's a biased coin toss. How am I supposed to find this out here? Otherwise, it would be E[X]=0.5

I do not believe that you are given a problem like this with absolutely no background. Maybe it was not covered in detail in class, but I would bet it is in your textbook, somewhere. If it is not in your textbook and not in your course notes, it is certainly widely available in other books or on line. I think you should make the effort to find it before asking us to do it for you.

RGV

XodoX said:

## Homework Statement

A biased coin is tossed ten times. Suppose that the probability of getting heads on a single toss is p. Let X be the number of heads obtained.
a)Give an algebraic formula for the probability mass function of X.
b) What do you think E[X] should be.

## The Attempt at a Solution

No idea. The probability mass function for a discrete random variable is:

pX(x)=Pr(X=x) for all x

As Ray pointed out, this problem is a very basic one.
However, some points for you (hope you know little probability):
a)
1) think of what is the probability that head is obtained once;
2) then, what is the probability that it is obtained X-times in 10 tosses? You will obtain pmf.

b) read carefully, how the mean value is defined for discrete distributions. It is very straightforward even to guess the result.

80past2 said:
Since they don't tell what you what the bias is, you can kind of ignore it. All you know is that the probability is p. What if you tossed a fair coin 10 times? What is the probability of getting x heads?

You are referring to be? I already answered that. It's 0.5. There are only 2 possible outcomes. So the probability is 0.5. Each time there's a 0.5 probability. That's all I can think of.
camillio said:
As Ray pointed out, this problem is a very basic one.
However, some points for you (hope you know little probability):
a)
1) think of what is the probability that head is obtained once;
2) then, what is the probability that it is obtained X-times in 10 tosses? You will obtain pmf.

b) read carefully, how the mean value is defined for discrete distributions. It is very straightforward even to guess the result.

It's obviously not "very basic" to me, since I would not have asked then. I don't know what you mean. This dosen't lead me to the answer. My best guess is pX(10)=0.5
I still don't know how to do b. E[X] is the expected value. 5 heads and 5 tails.

The probability of getting one head is p, right? What's the probability of getting two heads? Write out the different possibilities, of all the coin tosses. i.e. you can get two heads, a head or tail, a tail or head, or two heads. What are the associated probabilities for each of these conditions?

XodoX said:
You are referring to be? I already answered that. It's 0.5. There are only 2 possible outcomes. So the probability is 0.5. Each time there's a 0.5 probability. That's all I can think of.

It's obviously not "very basic" to me, since I would not have asked then. I don't know what you mean. This dosen't lead me to the answer. My best guess is pX(10)=0.5
I still don't know how to do b. E[X] is the expected value. 5 heads and 5 tails.

NO, NO, NO! You were *told* the probability of heads is p per toss. Do you still say that the probability of heads in one toss is 0.5? You were told it is not! Suppose I have a really biased coin with probability 0.999 of heads in each toss (and probability 0.001 of tails). Of course, in each trial there are only two outcomes---heads or tails---but that does not mean they are equally likely. They are not: P{heads} = 0.999 and P{tails} = 0.001.

I cannot figure out whether or not you are reading your textbook or going to class and taking notes. Something is not getting through to you. You are missing the very most basic ideas, and I have no idea how to help you.

RGV

XodoX said:
It's obviously not "very basic" to me, since I would not have asked then. I don't know what you mean. This dosen't lead me to the answer. My best guess is pX(10)=0.5
I still don't know how to do b. E[X] is the expected value. 5 heads and 5 tails.

XodoX, I did not intend to offend you. Please read twice what the problem is. You have (possibly biased) coin, i.e., Pr(one head in one toss) = p. What is then the probability of one tail in one toss?

Now think. You have 10 tosses and say you are calculating probability that you obtain right ONE head. This means 1 head AND 9 tails. Answer the following questions:
1) What is the probability that first 1 hat and then 9 tails occur?
2) How many events 1H + 9T may happen? Remind, you can get that head in ANY toss, so you will use basic combinatorics.

Then what rule applies to 2H + 8T. Very probably, you will get an idea what the final answer is.

80past2 said:
The probability of getting one head is p, right? What's the probability of getting two heads? Write out the different possibilities, of all the coin tosses. i.e. you can get two heads, a head or tail, a tail or head, or two heads. What are the associated probabilities for each of these conditions?

It's only a coin with 2 sides. Head and Tail have to have the same probability if I ignore it's biased. Each has a probability of 1/2. Two whatever is then 0.25, and one T or H is 0.5.

XodoX said:
It's only a coin with 2 sides. Head and Tail have to have the same probability if I ignore it's biased. Each has a probability of 1/2. Two whatever is then 0.25, and one T or H is 0.5.

You are not allowed to ignore the bias: you were told explicitly that the coin is biased. Go back to the example I gave you, where P{heads} = 0.999 and P{tails} = 0.001. In two tosses of this coin, what is the probability of two heads? of two tails? of one head and one tail?

You seem to be hung up on the false notion that because there are only two outcomes, they must be equally likely. NOT TRUE!

In fact, even for real coins in the real world there is evidence of bias. See, eg, the articles
http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition
and
http://www.statistik.lmu.de/~helmut/Texte/Euro.pdf .
As this last article points, some of the Euro coins in different countries have different amounts of bias; it also cites experimental work by Diaconis and others, pointing out the existence of bias in some actual coin-tossing situations.

RGV

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## What is a probability mass function (PMF) for a coin toss?

A probability mass function (PMF) for a coin toss is a mathematical function that assigns probabilities to each possible outcome of a coin toss. It is used to describe the distribution of a random variable, which in this case is the outcome of a coin toss.

## What are the possible outcomes of a coin toss?

The possible outcomes of a coin toss are two: heads or tails. These are mutually exclusive outcomes, meaning that only one of them can occur at a time.

## How is a probability mass function (PMF) for a coin toss calculated?

A probability mass function (PMF) for a coin toss is calculated by dividing the number of desired outcomes by the total number of possible outcomes. For example, if we want to calculate the PMF for getting heads on a fair coin toss, we would divide 1 (desired outcome) by 2 (total possible outcomes), giving us a PMF of 0.5.

## What is the difference between a probability mass function (PMF) and a probability density function (PDF)?

A probability mass function (PMF) is used for discrete random variables, such as the outcome of a coin toss. It assigns probabilities to each possible outcome. On the other hand, a probability density function (PDF) is used for continuous random variables, such as the weight of a person. It describes the probability of a random variable falling within a certain range of values.

## What is the importance of a probability mass function (PMF) for a coin toss?

A probability mass function (PMF) for a coin toss is important because it allows us to understand and predict the likelihood of different outcomes. It is also a fundamental concept in probability and statistics and is used in many real-world applications, such as gambling, finance, and risk analysis.

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