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# Probability S1 help watch

1. I have a work sheet (attached) but not sure if my answers are anywhere near right, so here they are for the first few questions. Would appreciate it if you could check whether they are correct so I know whether I am on the right track or not.

1) a) 0.3
b) 0.9
c) 0.56

2) a) 0.6
b) 0.45
c) 0.6

https://imgur.com/a/o8wa8
2. Q1

a) you are right

b) you cannot in general just add P(A) and P(B)... that would be counting the overlap section twice

c) think of it as "people who like apples & do not like bananas"
3. (Original post by JacobPowell)
I have a work sheet (attached) but not sure if my answers are anywhere near right, so here they are for the first few questions. Would appreciate it if you could check whether they are correct so I know whether I am on the right track or not.

1) a) 0.3
b) 0.9
c) 0.56

2) a) 0.6
b) 0.45
c) 0.6

https://imgur.com/a/o8wa8

1) a) 0.3 - fine
b) 0.9 - wrong, you forgot to do one more thing. P(A)+P(B) means you count the intersection TWICE but you want it only once.
c) 0.56 - wrong, draw a diagram and shade in A and B'. What is the region where the two intersect?

2) a) 0.6 - fine
b) 0.45 - wrong, draw a diagram again, except the two events don't intersect since they are mutually exclusive. Then you find that the intersection between their complements is 1-P(A)-P(B)
c) 0.6 - Wrong. Again, draw a diagram, the 'or' means either one shaded region (A' ) or the other (B' ). How much of the whole thing do you have shaded? EDIT: Or you can just use the formula P(A)+P(B)-P(AnB) with appropriate events, and use your answer to part b.
4. (Original post by RDKGames)
1) a) 0.3 - fine
b) 0.9 - wrong, you forgot to do one more thing. P(A)+P(B) means you count the intersection TWICE but you want it only once.
c) 0.56 - wrong, draw a diagram and shade in A and B'. What is the region where the two intersect?

2) a) 0.6 - fine
b) 0.45 - wrong, draw a diagram again, except the two events don't intersect since they are mutually exclusive. Then you find that the intersection between their complements is 1-P(A)-P(B)
c) 0.6 - Wrong. Again, draw a diagram, the 'or' means either one shaded region (A' ) or the other (B' ). How much of the whole thing do you have shaded? EDIT: Or you can just use the formula P(A)+P(B)-P(AnB) with appropriate events, and use your answer to part b.
Thanks for the help.

So would 1)b) be 0.7 + 0.2 - (0.7 * 0.2) = 0.76.
1)c) be 0.7 * 0.2 = 0.14 as it’s finding the intersection of A and B.

2)b) P(A’nB’) is not the intersection of A or B so does it just equal 0?
5. (Original post by JacobPowell)
Thanks for the help.

So would 1)b) be 0.7 + 0.2 - (0.7 * 0.2) = 0.76.
1)c) be 0.7 * 0.2 = 0.14 as it’s finding the intersection of A and B.
The probability of the intersection of events is NOT given by their product.

2)b) P(A’nB’) is not the intersection of A or B so does it just equal 0?
Not 0, try again. As I said, draw a diagram. Shade in A' with one colour, then shade in B' with a different colour. What is the region where the colours intersect? So what it's probability?
6. (Original post by RDKGames)
Not 0, try again. As I said, draw a diagram. Shade in A' with one colour, then shade in B' with a different colour. What is the region where the colours intersect? So what it's probability?
Right, thanks...let’s try again

1)b) 0.8
c) 0.1??

2)b) 1?
c) 0.4
7. (Original post by JacobPowell)
Right, thanks...let’s try again

1)b) 0.8
c) 0.1??

2)b) 1?
c) 0.4
1)b) 0.8 - correct.
c) 0.1?? - not quite. Can you show a diagram you drew? Or how you got that?

2)b) 1?
c) 0.4

Swap the answers around for Q2 here then you'll be right. If you don't understand why, then y'know, say why you disagree.

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