Challenging probability question HELP

Q)
The independent random variables

X

and

Y

have probability distributions:

P(X=x) = \frac{1}{8}, \ x= 1, 2, 3, 4, 5, 6, 7, 8

P(Y=y) = \frac{1}{y}, \ y= 2, 3, 6

Find

P(X > Y)

I'm really unsure of what to do with the information they've given me, any help is appreciated. Thanks

Form a list of the pairs for which X>Y and add up the probabilities. It may be quicker to do
1 - P(X<=Y).

Original post by ghall7

Q)
The independent random variables

X

and

Y

have probability distributions:

P(X=x) = \frac{1}{8}, \ x= 1, 2, 3, 4, 5, 6, 7, 8

P(Y=y) = \frac{1}{y}, \ y= 2, 3, 6

Find

P(X > Y)

I'm really unsure of what to do with the information they've given me, any help is appreciated. Thanks

OP

I still don't understand what to do with that :s-smilie:

:s-smilie:

.
Sorry

Original post by mqb2766

Form a list of the pairs for which X>Y and add up the probabilities. It may be quicker to do
1 - P(X<=Y).

Which bit don't you understand, can you list all (x,y) pairs for which x>y?

Original post by ghall7

I still don't understand what to do with that :s-smilie:

:s-smilie:

.
Sorry

OP

(x, y)

______________________________

(3, 2)

(4, 2)

(4, 3)

(5, 2)

(5, 3)

(6, 2)

(6, 3)

(7, 2)

(7, 3)

(7, 6)

(8, 2)

(8, 3)

(8, 6)

You mean like this?
I'm not sure how to work out the probabilities of those pairs occurring though, since the events are independent.

Original post by mqb2766

Which bit don't you understand, can you list all (x,y) pairs for which x>y?

(edited 4 years ago)

Its a basic/key result.
When the random variables are independent, how do you calculate the joint?
p(x&y) = p(x)???p(y)

Original post by ghall7

(x, y)

______________________________

(3, 2)

(4, 2)

(4, 3)

(5, 2)

(5, 3)

(6, 2)

(6, 3)

(7, 2)

(7, 3)

(7, 6)

(8, 2)

(8, 3)

(8, 6)

You mean like this?
I'm not sure how to work out the probabilities of those pairs occurring though, since the events are independent.

Study Forum Helper

Original post by ghall7

(x, y)

______________________________

(3, 2)

(4, 2)

(4, 3)

(5, 2)

(5, 3)

(6, 2)

(6, 3)

(7, 2)

(7, 3)

(7, 6)

(8, 2)

(8, 3)

(8, 6)

You mean like this?
I'm not sure how to work out the probabilities of those pairs occurring though, since the events are independent.

But that's just it... because they're independent, it means that

P(\{ X = 3 \} \cap \{ Y = 2\}) = P(X = 3) \cdot P(Y=2)

for the top case.

OP

When you say the joint of the two events, is that the same as saying the intersection of the two events or (x AND y)?

(x ∩ y)

Original post by mqb2766

Its a basic/key result.
When the random variables are independent, how do you calculate the joint?
p(x&y) = p(x)???p(y)

(edited 4 years ago)

That both events occur
p(x,y) = p(x|y)p(y) = ...
The independent part is a long winded (but correct) of relating the condition p(x|y) = p(x), so the probability of x, does not depend on y.

Original post by ghall7

When you say the joint of the two events, is that the same as saying the intersection of the two events or 'and'?

(x ∩ y)

OP

I understand what you're saying, did you get

P(\{ X = 3 \} \cap \{ Y = 2\}) = P(X = 3) \cdot P(Y=2)

from the fact that when two events are independent,

P(y|x) = P(y)

?

Original post by RDKGames

But that's just it... because they're independent, it means that

P(\{ X = 3 \} \cap \{ Y = 2\}) = P(X = 3) \cdot P(Y=2)

for the top case.

OP

Okay, I understand a bit more now.

I got 0.625 as my answer.

Original post by mqb2766

That both events occur
p(x,y) = p(x|y)p(y) = ...
The independent part is a long winded (but correct) of relating the condition p(x|y) = p(x), so the probability of x, does not depend on y.

I've not calcuated it explicitly, but it seems sensible.
That is correct about independent, the joint is just the product of the individual probabilities, or equivalently, the conditional is just the individual.

Original post by ghall7

Okay, I understand a bit more now.

I got 0.625 as my answer.

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