# Need your help in probability problems? Thanks :)

Apr 29, 2018

See below

#### Explanation:

P(x)=$\frac{8}{12} = \frac{2}{3}$

P(Y)=$\frac{5}{12}$

P(X nn Y)$= \frac{2}{12} = \frac{1}{6}$ the numbers 1 and 3 are in both events.

P(X uu Y)$= \frac{11}{12}$ the number 12 is not in either event

The sample space would be the numbers 1 to 12

The events are dependent as 1 and 3 are in both events

Given event A has occurred P(even)=$\frac{2}{8} = \frac{1}{4}$
Sorry don't know what P(X/Y) means

Apr 29, 2018

### What is $\text{P} \left(X\right)$?

The probability of event $X$ occurring is

$\text{P"(X)="the number of elements in X"/"the number of elements in S}$

Since there are 12 possible rolls, $S$ has 12 elements. The event $X$ contains 8 of those 12. So, $\text{P} \left(X\right)$ is

$\text{P} \left(X\right) = \frac{8}{12}$

$\textcolor{w h i t e}{\text{P} \left(X\right)} = \frac{2}{3}$

### What is $\text{P} \left(Y\right)$?

Similarly, $Y$ contains 5 of the 12 possible rolls, so

$\text{P} \left(Y\right) = \frac{5}{12}$

### What is $\text{P} \left(X \cap Y\right)$?

The event $X \cap Y$ is the set of elements that are in both $X$ and $Y$. By observation, we see that $X \cap Y = \left\{1 , 3\right\} .$ Thus, the size of $X \cap Y$ is 2, and the probability of $X \cap Y$ is

$\text{P} \left(X \cap Y\right) = \frac{2}{12}$

$\textcolor{w h i t e}{\text{P} \left(X \cap Y\right)} = \frac{1}{6}$

### What is $\text{P} \left(X \cup Y\right)$?

The event $X \cup Y$ is the set of elements that are in $X$ or $Y$. If an element appears in at least one of $X$ or $Y$, then it is in $X \cup Y$.

By observation, we see $X \cup Y = \left\{1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11\right\}$. (The only element that does not appear in either $X$ or $Y$ is $\left\{12\right\}$.) Thus the probability of $X \cup Y$ is

$\text{P} \left(X \cup Y\right) = \frac{11}{12}$

### Write down the sample space $S$ of possible outcomes.

$S$ is the event that anything happens. It contains all the elements that we could possibly observe, which makes "P"(S)=100%. For a standard 12-sided die, the sample space of a single roll is

$S = \left\{1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12\right\}$

### Are $X$ and $Y$ independent? Explain.

Two events $X$ and $Y$ are independent if

$\text{P"(XnnY) = "P"(X) xx "P} \left(Y\right)$

From above, we know $\text{P} \left(X \cap Y\right) = \frac{1}{6}$. We also know $\text{P} \left(X\right) = \frac{2}{3}$ and $\text{P} \left(Y\right) = \frac{5}{12}$. We plug these values into the equation to check:

$\frac{1}{6} \stackrel{\text{? }}{=} \frac{2}{3} \times \frac{5}{12} = \frac{1 \times 5}{3 \times 6} = \frac{5}{18} \ne \frac{1}{6}$

Since $\text{P"(XnnY) != "P"(X) xx "P} \left(Y\right)$, we know $X$ and $Y$ are not independent.

### What is the probability that you rolled an even number, given that event $A$ has occurred?

There is no event $A$ defined in the question. Assuming they mean event $X$, we seek the ratio

$\text{P"("even"|X)= "the number of even numbers in X"/"the size of X}$

Event $X$ has 2 even numbers in it: $\left\{4 , 8\right\}$. So the probability of rolling an even number, given that $X$ has occurred, is:

"P"("even"|X)= 2/8

color(white)("P"("even"|X))= 1/4

### What is $\text{P} \left(X | Y\right)$?

Similarly, $\text{P} \left(X | Y\right)$ is the probability of rolling a number that's also in $X$, given that it's known to be in $Y$.

"P"(X|Y) =("the size of X "nn" Y")/"the size of Y"= ("P"(X nn Y))/("P"(Y))

$\textcolor{w h i t e}{\text{P} \left(X | Y\right)} = \frac{2}{5}$