# Venn Diagram Unknown Values

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#1
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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#2
mmmmmmm
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3 weeks ago
#3
What do you understand about independence for joint probabilities?
Last edited by mqb2766; 3 weeks ago
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#4
(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.
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3 weeks ago
#5
(Original post by 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.
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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#6
(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) ?
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3 weeks ago
#7
(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/prob...dependent.html
Last edited by mqb2766; 3 weeks ago
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#8
(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/prob...dependent.html
No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(B)
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#9
(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/prob...dependent.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)
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3 weeks ago
#10
(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?
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3 weeks ago
#11
(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?
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#12
(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/prob...dependent.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?
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#13
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?
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3 weeks ago
#14
It's not an equation. You've not put an = ?
Leave it factorized once you've made the equation.
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#15
(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
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3 weeks ago
#16
(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.
Last edited by mqb2766; 3 weeks ago
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#17
(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) ?
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3 weeks ago
#18
(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.
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#19
P (A n C) = z
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3 weeks ago
#20
(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/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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