# The joint probability - two coins

• docnet
In summary, the joint probability of getting heads on both coins is 1/4 or 25%. It is calculated by multiplying the individual probabilities of each event. Joint probability and conditional probability differ in that joint probability looks at the likelihood of two events occurring together, while conditional probability looks at the likelihood of one event occurring given another event has already occurred. The joint probability cannot be greater than 1 and is commonly used in fields such as statistics, probability, and data analysis to make predictions and solve problems.
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Homework Statement
.
Relevant Equations
.
(a)

The probability mass function of X and Y:

$$p_{X,Y}(1,1)= \frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$$
$$p_{X,Y}(1,0)=\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$$
$$p_{X,Y}(0,1)= \frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$$
$$p_{X,Y}(0,0)= \frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$$

or

$$p_{X,Y}(a,b)= \begin{cases} \frac{3}{8} & (a,b)=(1,1) \quad \text{or}\quad(a,b)=(0,1) \\ \frac{1}{8} & (a,b)=(1,0) \quad \text{or}\quad(a,b)=(0,0) \end{cases}$$

(b)

The marginal distribution of X is

$$p_X(1)=p_{X,Y}(1,0)+p_{X,Y}(1,1)=\frac{1}{2}\frac{3}{4}+\frac{1}{2}\frac{1}{4}=\frac{1}{2}$$
$$p_X(0)=p_{X,Y}(0,0)+p_{X,Y}(0,1)=\frac{1}{2}\frac{1}{4}+\frac{1}{2}\frac{3}{4}=\frac{1}{2}$$

The marginal distribution of Y is

$$p_Y(1)=p_{X,Y}(0,1)+p_{X,Y}(1,1)=\frac{1}{2}\frac{3}{4}+\frac{1}{2}\frac{1}{4}=\frac{3}{4}$$
$$p_Y(0)=p_{X,Y}(0,0)+p_{X,Y}(1,0)=\frac{1}{2}\frac{1}{4}+\frac{1}{2}\frac{3}{4}=\frac{1}{4}$$

(c)

The expected value of Y is

$$E(Y)=1\cdot p_Y(1)+0\cdot p_Y(0)=\frac{3}{4}$$

(d)

$$E((X-0.5)\times Y)=\sum\limits_{(a,b)}(a-.5)bp_{X,Y}(a,b)$$
$$(1-0.5)(1)p_{X,Y}(1,1)+(1-0.5)(0)p_{X,Y}(1,0)+$$
$$(0-0.5)(1)p_{X,Y}(0,1)+(0-0.5)(0)p_{X,Y}(0,0)=$$
$$\frac{1}{2}\cdot\frac{3}{8}+0-\frac{1}{2}\cdot \frac{3}{8}+0=0$$

docnet said:
Homework Statement:: .
Relevant Equations:: .

View attachment 293161(a)

The probability mass function of X and Y:

$$p_{X,Y}(1,1)= \frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$$
$$p_{X,Y}(1,0)=\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$$
$$p_{X,Y}(0,1)= \frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$$
$$p_{X,Y}(0,0)= \frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$$
That's not what I get. If ##X = 1##, then we toss the unbiased coin (coin 1) again.

PeroK said:
That's not what I get. If ##X = 1##, then we toss the unbiased coin (coin 1) again.

yes! the problem I solved isn't same as the problem in the prompt :(

but don't worry, i think i know what the problem is asking, and how to solve it

OK so here's what I have

(a)

$$P_{X,Y}(1,1)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}$$
$$P_{X,Y}(1,0)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}$$
$$P_{X,Y}(0,1)=\frac{1}{2}\frac{3}{4}=\frac{3}{8}$$
$$P_{X,Y}(1,1)=\frac{1}{2}\frac{1}{4}=\frac{1}{8}$$(b)

$$P_X(1)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
$$P_X(0)=\frac{3}{8}+\frac{1}{8}=\frac{1}{2}$$
$$P_Y(1)=\frac{3}{8}+\frac{1}{4}=\frac{5}{8}$$
$$P_Y(0)=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$$

(c)

$$E(Y)=\frac{5}{8}+0\cdot \frac{3}{8}=\frac{5}{8}$$

(d)

$$E((X-.5)\times Y)=(\frac{1}{2}\frac{1}{2}-\frac{1}{2}\frac{1}{2})\frac{5}{8}=0$$

It looks right apart from the answer to d). Note that in general ##E(XY) \neq E(X)E(Y)##. (Another simple but erroneous calculation on your part!) But, in general, ##E(X + Y) = E(X) + E(Y)##. That gives you two ways to do part d): go through the four options, or use the expected value of a sum.

@PeroK thank you :) is this what you mean?

using the formula:
$$E((X-.5)Y)=\frac{1}{8}-\frac{3}{16}=-\frac{1}{16}$$ using linearity and the definition of covariance:

$$E((X-.5)Y)=E(XY)-\frac{5}{16}$$

$$E(XY)=cov(X,Y)+E(X)E(Y)=-\frac{1}{16}+\frac{5}{16}$$

$$E((X-.5) Y)=-\frac{1}{16}$$

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PeroK
Yes, it's good to see you do it both ways and get the same answer!

Note: the quick and valid way to get ##E(XY)## is to note that ##XY## is only non-zero when ##X = Y = 1##, which happens with probability ##1/4##. Hence ##E(XY) = \frac 1 4##.

docnet

## 1. What is the joint probability of getting two heads when flipping two coins?

The joint probability of getting two heads when flipping two coins is 1/4 or 25%. This means that out of all the possible outcomes (HH, HT, TH, TT), there is only one way to get two heads.

## 2. How is the joint probability calculated for two coins?

The joint probability for two coins is calculated by multiplying the individual probabilities of each event. For example, the probability of getting heads on the first coin is 1/2, and the probability of getting heads on the second coin is also 1/2. Multiplying 1/2 by 1/2 gives us a joint probability of 1/4.

## 3. What is the difference between joint probability and conditional probability?

Joint probability is the likelihood of two events occurring together, while conditional probability is the likelihood of one event occurring given that another event has already occurred. In the context of two coins, joint probability would be the likelihood of getting two heads, while conditional probability would be the likelihood of getting a head on the second coin given that the first coin landed on heads.

## 4. Can the joint probability of two coins ever be greater than 1?

No, the joint probability of two coins can never be greater than 1. This is because the sum of all possible outcomes must equal 1, and the joint probability of two events occurring together is always less than or equal to the probability of each event occurring individually.

## 5. How can joint probability be used in real-life situations?

Joint probability can be used in many real-life situations, such as in genetics to calculate the likelihood of inheriting certain traits from both parents, in finance to determine the probability of two stocks performing well together, and in medicine to assess the effectiveness of a treatment when two conditions are present.

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