Venn Diagram

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kparikh
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#1
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#1
The figure that I have attached below shows a Venn diagram of the two events A and B.

P(A and not B) = 0.4, P(B and not A) = 0.1, P(A and B) = x

The question asks:

a. Given that A and B are independent, write down an equation
in x.

b. Solve this equation and hence find the possible pairs of values
for x and y.

My response:

I believe that the equation would be x^2 - 0.5x + 0.04, since A = 0.4-x and B = 0.1 - x

So A ∩ B = P(A) * P(B)

Therefore (0.4-x)(0.1-x) = x^2 - 0.5x + 0.04

For question b, my understanding is to solve (0.4-x)(0.1-x) = 0, to get x

Therefore x would be 0.4 and 0.1

Since we know the values for x, I assume that y could then be founding by solving:

1-(0.4+0.4+0.1) = 0.1
1-(0.4+0.1+0.1) = 0.4

My only query is that I feel that my response is to short for the given marks of the question b (5 marks).

Would this approach to both question a and b be accurate?

Thank You for any guidance you may have to share.
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kparikh
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#2
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#2
Name:  Screen Shot 2022-01-15 at 20.03.03.png
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mqb2766
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#3
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#3
Youre thinking along the right lines, but think about what region the probabilities 0.4 and 0.1 refer to and what is p(A) and p(B).

Then if they're independent, the equation (part a) must be
p(AnB) = p(A)*p(B)
Youve not included the left hand side and you've stated an expression, not an equation.
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kparikh
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#4
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#4
would it be x = x^2 - 0.5x + 0.04

since P(AnB) represents the intersection which is x

therefore 0 = x^2 - 1.5x + 0.04

If this is accurate, how would I proceed in answering question b then?
Thank You so much for the guidance
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mqb2766
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#5
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#5
(Original post by kparikh)
would it be x = x^2 - 0.5x + 0.04

since P(AnB) represents the intersection which is x

therefore 0 = x^2 - 1.5x + 0.04

If this is accurate, how would I proceed in answering question b then?
Thank You so much for the guidance
Agree with the left hand side of the bold, but the right hand side is wrong, because of the reasons in the previous post. What regions do 0.4 and 0.1 correspond to, so what is p(A) and p(B)? Your expressions are not correct so neither is the product of them.

Edit - it would be worth noting that your original solutions for the joint wouldn't really make sense as they would correspond to one of the probabilities p() = 0.
Last edited by mqb2766; 3 months ago
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kparikh
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#6
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#6
I assume that P(A) = (x-0.4) and P(B) = (x-0.1)

My apologies, I am unable to see what else it could correspond to.

Thank You.
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mqb2766
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#7
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#7
(Original post by kparikh)
I assume that P(A) = (x-0.4) and P(B) = (x-0.1)

My apologies, I am unable to see what else it could correspond to.

Thank You.
This is core gcse stuff
https://www.bbc.co.uk/bitesize/guide...wxs/revision/5
How do you find each circle, p(A) and p(B), from each of the parts? You're close but ...
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kparikh
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#8
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#8
would 0.5 + x = P(A) + P(B)?

Of not, would it be possible for you to share the values of P(A) and P(B) so i can work backwards.

Thank You.
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mqb2766
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#9
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#9
(Original post by kparikh)
would 0.5 + x = P(A) + P(B)?

Of not, would it be possible for you to share the values of P(A) and P(B) so i can work backwards.

Thank You.
You really should know this but
p(A) = 0.4 + x
and similarly for p(B).

Maths way
p(A) = p(A n B) + p(A n B')
Venn way, the circle, p(A), is made from the two parts 0.4 and x.
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kparikh
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#10
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#10
Ah, ok thank you.

so i would use this to then to achieve the equation x = (x+0.4)(x+0.1)

this would then give, x^2 + 0.5x + 0.04 = x

so x^2 - 0.5x + 0.04 = 0

doesn't this just provide me with the equation that I was initially trying to solve

so x = 0.4 and 0.1?
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mqb2766
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#11
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#11
(Original post by kparikh)
Ah, ok thank you.

so i would use this to then to achieve the equation x = (x+0.4)(x+0.1)

this would then give, x^2 + 0.5x + 0.04 = x

so x^2 - 0.5x + 0.04 = 0

doesn't this just provide me with the equation that I was initially trying to solve

so x = 0.4 and 0.1?
Your expression which you tried to "solve" did give the same answers when you set it equal to zero. However, you forgot to mention why you set it equal to zero and what the solutions meant. It was also the wrong expressoin for what you were trying to form etc.

But as noted in a previous post according to the OP, those solutions corresponded to
x = 0.4 => p(A) = 0, p(B) = -0.3
x = 0.1 => p(A) = 0.3, p(B) = 0
You realize this is just wrong?

If you want to verify the new work, sub the x values back into p(A) and p(B) and verify the independence property holds.
Last edited by mqb2766; 3 months ago
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kparikh
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#12
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#12
How would I then go about finding x and y, since x = 0.4 and 0.1 provides inaccurate solutions

That You so much for your help
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mqb2766
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#13
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#13
(Original post by kparikh)
How would I then go about finding x and y, since x = 0.4 and 0.1 provides inaccurate solutions

That You so much for your help
Not sure what you mean. The p(A) and p(B) I was talking about in the previous post were for the incorrect expressions in the OP. I was trying to illustrate the point that the original solutions were meaningless in terms of how the equation was set up. The values for x in the OP and #10 may be the same, but the original expressions are different, so the values are interpreted differently.

For your work in #10, what are p(A) and p(B) for those values of x and does the indepence property hold in each case?
Last edited by mqb2766; 3 months ago
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kparikh
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#14
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#14
So if x = 0.4 then y = 1 - (0.4 + 0.4 + 0.1) = 0.1

if x = 0.1 then y = 1 - (0.4 + 0.1 +0.1) = 0.4

Would this be accurate?

Thank You
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mqb2766
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#15
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#15
(Original post by kparikh)
So if x = 0.4 then y = 1 - (0.4 + 0.4 + 0.1) = 0.1

if x = 0.1 then y = 1 - (0.4 + 0.1 +0.1) = 0.4

Would this be accurate?

Thank You
Edit y is on the venn.

But for verifying the solutions I meant
x = 0.4 => p(A) = ..., p(B) = ..., so p(A)*p(B) = ... and does it equal x?
Similarly for 0.1.
Last edited by mqb2766; 3 months ago
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kparikh
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#16
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#16
For 0.4

(0.4+0.4)*(0.1+0.4) = 0.4 so the case holds for x = 0.4

For 0.1

(0.1+0.4)(0.1+0.1) = 0.1 so the case also holds for x = 0.1

Does this mean that for question b, the answer would be:

x = 0.4, y = 0.1
x = 0.1, y = 0.4

Thank you for explaining the process to me.
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mqb2766
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#17
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#17
(Original post by kparikh)
For 0.4

(0.4+0.4)*(0.1+0.4) = 0.4 so the case holds for x = 0.4

For 0.1

(0.1+0.4)(0.1+0.1) = 0.1 so the case also holds for x = 0.1

Does this mean that for question b, the answer would be:

x = 0.4, y = 0.1
x = 0.1, y = 0.4

Thank you for explaining the process to me.
Where does the question come from?
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kparikh
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#18
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#18
The question is to solve the equation, x^2-0.5x+0.04 and hence find the possible pairs of values for x and y.
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Gaurang12
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#19
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#19
Hint :

 P(A) = P(A \cap B) + P(A \cap B^{\prime})

 P(B) = P(A \cap B) + P(A^{\prime} \cap B)

 P(A \cup B)^{\prime} = 1- P(A  \cup B) = y

IF A and B are independent it implies the following and these are independent as well :

 P(A \cap B^{\prime}) = P(A) \cdot P(B^{\prime})

 P(A^{\prime} \cap B) = P(A^{\prime}) \cdot P(B)
Last edited by Gaurang12; 3 months ago
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mqb2766
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#20
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#20
(Original post by kparikh)
The question is to solve the equation, x^2-0.5x+0.04 and hence find the possible pairs of values for x and y.
Sure I was asking where it came from? Is it an exam question / what level ...
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