username4506446
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I’m stuck on this question. For part a) how would I draw the diagram. I know if they are mutually exclusive they don’t cross over. Thanks
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old_engineer
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(Original post by Anonymous_4657)
I’m stuck on this question. For part a) how would I draw the diagram. I know if they are mutually exclusive they don’t cross over. Thanks
What you have said describes the Venn diagram completely, so you can just draw it. You could make one blob bigger than the other to reflect the relative probabilities suggested in the question, but beyond that Venn diagrams are not usually drawn to scale.
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(Original post by old_engineer)
What you have said describes the Venn diagram completely, so you can just draw it. You could make one blob bigger than the other to reflect the relative probabilities suggested in the question, but beyond that Venn diagrams are not usually drawn to scale.
I don’t understand the bit which says p is not equal to zero.
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(Original post by old_engineer)
What you have said describes the Venn diagram completely, so you can just draw it. You could make one blob bigger than the other to reflect the relative probabilities suggested in the question, but beyond that Venn diagrams are not usually drawn to scale.
How do I do part (i) (b)
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ThiagoBrigido
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Total probability equals 1. Therefore A=3B;
A+B=1
(3B)+B=1
4B=1
B=0.25
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old_engineer
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(Original post by ThiagoBrigido)
Total probability equals 1. Therefore A=3B;
A+B=1
(3B)+B=1
4B=1
B=0.25
That's not quite it. The two blobs do not necessarily occupy the whole sample space, and it is the whole sample space that sums to 1. You have correctly determined that P(A) + P(B) = 4P(B). The constraint we now have is 4P(B) <= 1. This establishes the possible values of P(B).
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(Original post by old_engineer)
That's not quite it. The two blobs do not necessarily occupy the whole sample space, and it is the whole sample space that sums to 1. You have correctly determined that P(A) + P(B) = 4P(B). The constraint we now have is 4P(B) <= 1. This establishes the possible values of P(B).
Can you do part ii) a) and b) please.
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old_engineer
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(Original post by Anonymous_4657)
Can you do part ii) a) and b) please.
For part (ii)a, if C and D are independent, then P(C given D) = P(C). That's a basic tenet of conditional probability, which you should have come across if you have covered this topic.

For part (ii)b, I would recommend a version of the Venn diagram labelled as shown below.

Name:  image0.jpeg
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From there, you can start with q + r + s + t =1. Then use the information given in the question to tease out the values of q, r, s and t. The question asks for the value of P(C), which is given by q + r.
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(Original post by old_engineer)
For part (ii)a, if C and D are independent, then P(C given D) = P(C). That's a basic tenet of conditional probability, which you should have come across if you have covered this topic.

For part (ii)b, I would recommend a version of the Venn diagram labelled as shown below.

Name:  image0.jpeg
Views: 14
Size:  79.6 KB

From there, you can start with q + r + s + t =1. Then use the information given in the question to tease out the values of q, r, s and t. The question asks for the value of P(C), which is given by q + r.
I don’t understand??
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old_engineer
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(Original post by Anonymous_4657)
I don’t understand??
Then I think perhaps you need to try some simpler questions involving Venn diagrams and conditional probability, referring back to your textbook or lecture notes as needed.
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