# Venn Diagram Unknown Values

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Hi,

Please can I have some help with this question.

I've found X = 0

But need help finding y and z please?

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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#3

What do you understand about independence for joint probabilities?

Last edited by mqb2766; 3 weeks ago

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(Original post by

What do you understand about independence for joint probabilities?

**mqb2766**)What do you understand about independence for joint probabilities?

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

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#5

(Original post by

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.

**M.Johnson2111**)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.

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(Original post by

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

**mqb2766**)Sure, but what does it mean about how p(A and C) can be calculated from p(A) and p(C)?

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#7

(Original post by

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

**M.Johnson2111**)I'm not sure? Is the P (A) = P (C) ?

Do you not have the definition in your notes, it's very common?

https://www.mathsisfun.com/data/prob...dependent.html

Last edited by mqb2766; 3 weeks ago

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(Original post by

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/prob...dependent.html

**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/prob...dependent.html

Thank you.

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

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**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/prob...dependent.html

Thank you.

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

Which would be:

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

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#10

(Original post by

No, I started the topic yesterday.

Thank you.

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

**M.Johnson2111**)No, I started the topic yesterday.

Thank you.

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

That's correct. Can you express that in terms of the two unknowns?

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**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/prob...dependent.html

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

For starting yesterday, it's a bit complex, but possible.

That's correct. Can you express that in terms of the two unknowns?

**mqb2766**)For starting yesterday, it's a bit complex, but possible.

That's correct. Can you express that in terms of the two unknowns?

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?

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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?

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?

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#14

It's not an equation. You've not put an = ?

Leave it factorized once you've made the equation.

Leave it factorized once you've made the equation.

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(Original post by

It's not an equation. You've not put an = ?

Leave it factorized once you've made the equation.

**mqb2766**)It's not an equation. You've not put an = ?

Leave it factorized once you've made the equation.

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#16

(Original post by

Z (z + 0.49 + y) = 0.039

**M.Johnson2111**)Z (z + 0.49 + y) = 0.039

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.

Last edited by mqb2766; 3 weeks ago

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(Original post by

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.

**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.

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

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#18

You need an equation in terms of y and z to solve.

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#20

(Original post by

P (A n C) = z

**M.Johnson2111**)P (A n C) = z

You need another equation as you have two variables to determine. Can you think what that is?

https://www.ck12.org/probability/ven...bility-ALG-II/

It's not explicitly stated in the question, but it's a common property for probabilities.

Last edited by mqb2766; 3 weeks ago

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