Question

Need your help in probability problems? Thanks :)

Answer 1

See below

Explanation:

P(x)=`8/12=2/3`

P(Y)=`5/12`

P(X nn Y)`=2/12=1/6` the numbers 1 and 3 are in both events.

P(X uu Y)`=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)=`2/8=1/4`
Sorry don't know what P(X/Y) means

Answer 2

What is `"P"(X)`?

The probability of event `X` occurring is

`"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, `"P"(X)` is

`"P"(X)=8/12`

`color(white)("P"(X)) = 2/3`

What is `"P"(Y)`?

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

`"P"(Y)=5/12`

What is `"P"(X nn Y)`?

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

`"P"(XnnY) = 2/12`

`color(white)("P"(XnnY)) = 1/6`

What is `"P"(XuuY)`?

The event `XuuY` 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 `XuuY`.

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

`"P"(XuuY) = 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={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}`

Are `X` and `Y` independent? Explain.

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

`"P"(XnnY) = "P"(X) xx "P"(Y)`

From above, we know `"P"(XnnY)=1/6`. We also know `"P"(X)=2/3` and `"P"(Y)=5/12`. We plug these values into the equation to check:

`1/6 stackrel"? "= 2/3 xx 5/12=(1xx5)/(3xx6)=5/18 != 1/6`

Since `"P"(XnnY) != "P"(X) xx "P"(Y)`, 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

`"P"("even"|X)= "the number of even numbers in X"/"the size of X"`

Event `X` has 2 even numbers in it: `{4, 8}`. 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 `"P"(X|Y)`?

Similarly, `"P"(X|Y)` 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))`

`color(white)("P"(X|Y)) = 2/5`