Venn Diagram Unknown Values

M.Johnson2111

Hi,

Please can I have some help with this question.

I've found X = 0

But need help finding y and z please?

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mmmmmmm

What do you understand about independence for joint probabilities?

(edited 3 years ago)

Reply 3

M.Johnson2111

Original post by mqb2766

What do you understand about independence for joint probabilities?

If two events are independent of each other then the occurrence of one doesn't affect the probability of the occurrence of the other.

In this case I think this means the probability of event A doesn't affect the probability of the occurrence of event C.

Reply 4

mqb2766

Original post by M.Johnson2111

Sure, but what does it mean about how p(A and C) can be calculated from p(A) and p(C)?

Reply 5

M.Johnson2111

Original post by mqb2766

Sure, but what does it mean about how p(A and C) can be calculated from p(A) and p(C)?

I'm not sure? Is the P (A) = P (C) ?

Reply 6

mqb2766

Original post by M.Johnson2111

I'm not sure? Is the P (A) = P (C) ?

No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/probability-events-independent.html

(edited 3 years ago)

Reply 7

M.Johnson2111

Original post by mqb2766

No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/probability-events-independent.html

No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(B)

Reply 8

M.Johnson2111

Original post by mqb2766

No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/probability-events-independent.html

No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(C)

Which would be:

(0.10 + y + z) x (0.39 + z)

Reply 9

mqb2766

Original post by M.Johnson2111

No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(B)

For starting yesterday, it's a bit complex, but possible.
That's correct. Can you express that in terms of the two unknowns?

Reply 10

mqb2766

Original post by M.Johnson2111

(0.10 + y + z) x (0.39 + z)

What is the last expression equal to (in terms of the unknown?

Reply 11

M.Johnson2111

Original post by mqb2766

No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/probability-events-independent.html

No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(C)

Which would be:

(0.10 + y + z) x (0.39 + z)

Original post by mqb2766

For starting yesterday, it's a bit complex, but possible.
That's correct. Can you express that in terms of the two unknowns?

By expanding the brackets I got:

Z^2 + 0.49z + zy + 0.039

This looks like a quadratic equation which I could solve but the zy confuses me. Would I take a factor of Z out of the expression?

Reply 12

M.Johnson2111

By expanding the brackets I got:

Z^2 + 0.49z + zy + 0.039

This looks like a quadratic equation which I could solve but the zy confuses me. Would I take a factor of Z out of the expression?

Reply 13

mqb2766

It's not an equation. You've not put an = ?
Leave it factorized once you've made the equation.

Reply 14

M.Johnson2111

Original post by mqb2766

It's not an equation. You've not put an = ?
Leave it factorized once you've made the equation.

Z (z + 0.49 + y) = 0.039

Reply 15

mqb2766

Original post by M.Johnson2111

Z (z + 0.49 + y) = 0.039

You can't just stick an = into an expression. You have
So P(A and C) = P(A) x P(C)
Which would be
? = (0.10 + y + z) (0.39 + z)

What is the variable representation of p(A and B)? That's what goes on the left hand side.

(edited 3 years ago)

Reply 16

M.Johnson2111

Original post by mqb2766

You can't just stick an = into an expression. You have
So P(A and C) = P(A) x P(C)
Which would be
? = (0.10 + y + z) x (0.39 + z)

What is the variable representation of p(A and B)? That's what goes on the left hand side.

By this do you mean:

P(A and C) = (0.10 + y + z) x (0.39 + z) ?

Reply 17

mqb2766

Original post by M.Johnson2111

By this do you mean:

P(A and C) = (0.10 + y + z) x (0.39 + z) ?

From the Venn diagram, which letter(s) or expression represents the p(A and C)?
You need an equation in terms of y and z to solve.

P (A n C) = z

Original post by M.Johnson2111

P (A n C) = z

Yes, so the equation is ... ?

You need another equation as you have two variables to determine. Can you think what that is?
https://www.ck12.org/probability/venn-diagrams/lesson/Probability-Using-a-Venn-Diagram-and-Conditional-Probability-ALG-II/
It's not explicitly stated in the question, but it's a common property for probabilities.