# Venn Diagram

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kparikh

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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.

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

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.

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

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

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

would it be

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

**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

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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#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.

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

(Original post by

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.

**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.

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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#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.

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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#9

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#9

(Original post by

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.

**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.

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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#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?

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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#11

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#11

(Original post by

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?

**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?

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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#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

That You so much for your help

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#13

(Original post by

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

**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

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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#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

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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#15

(Original post by

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

**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

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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#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.

(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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#17

(Original post by

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.

**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.

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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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#20

(Original post by

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.

**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.

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