(Original post by z_o_e)
Can you check my answers please question 16 A..
I can't decrease for some reason...
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 20062016 21:14
Last edited by 34908seikj; 20062016 at 21:18. 
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 20062016 21:20
(Original post by 34908seikj)
Decrease?
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 20062016 21:21

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 20062016 21:26
(Original post by 34908seikj)
dividing by a number less than one would increase the number, multiplying a number less than one would decrease the number!
If the decimal is bigger than 1st divide.
If it is smaller than 1 I multiply.
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 20062016 21:26
(Original post by z_o_e)
Heya, my end of year exam is coming up and I'm doing higher and I'm getting D's I'm currently in year 10 and I know I can do better if I revise. Is anyone else willing to help me with a few questions ?
I am a year 11 student and will be finishing my last exam soon. I used to struggle with maths a lot in year 9, but after practicing everyday by buying maths' revision books and really learning the techniques, then try to find questions related to those topics and working them out, was a huge help. I never really did many past papers but I went from an F in year 9 to an A* now in year11, through just a lot of practice. But try to print yourself a booklet of past papers, pretend it's like a work book and work through it. Ask your teachers to mark them. It would surprise you with how effective it's at making you familiar with the formatting of exam paper (this is important).
I wish you the best and do great in your GCSE. As I have done my GCSE (almost), the experience is daunting but be prepared and work hard. 
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 20062016 21:35
(Original post by z_o_e)
So if there's a decrease question.
If the decimal is bigger than 1st divide.
If it is smaller than 1 I multiply.
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 24062016 22:18
(Original post by 34908seikj)
Yes. But it does depend on the question.
For this do I multiply or add the values to find the answers.
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 24062016 23:10
(Original post by z_o_e)
HeyaaxFor this do I multiply or add the values to find the answers.Posted from TSR Mobile
How'd the exam go? 
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 24062016 23:13
(Original post by 34908seikj)
For part A, multiply. For part B, multiply then add.
How'd the exam go?
Terribly hard.
Lower bound and upper bound came up.
Locai.
Hard equations...
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 24062016 23:20
(Original post by z_o_e)
How do I know if it's multiply or add?
Terribly hard.
Lower bound and upper bound came up.
Locai.
Hard equations...
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(part b) At least one of them is at school = 1 attends AND one doesn't
0.4*0.7 = 0.28  probability of Bill attending WHEN Anna doesn't
0.6*0.3 = 0.18  probability of Anna attending WHEN Bill doesn't
You multiply because it's conditional probability  The probability of the second event depends on the first event.
Now we add 0.28 and 0.18 together to get the final answer of 0.46
We add because it's independent from the first event  no matter what the first event is the second event will not be affected, if that makes sense. 
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 25062016 09:22
(Original post by 34908seikj)
For part A, multiply. For part B, multiply then add.
How'd the exam go?
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 25062016 09:41
Instead of 0.4+0.4, it would be 0.4*0.4 = 0.16
Which you probably just wrote addition by mistake, given that you got the right answer!
Part B, You need, the probability he will win in game one * by the probability that he does not win game two.
Then you also need to probability he does not win a game * the probability that he wins a game. Then just add the two together to get your answer
P(W,NW)  Probability of winning then not winning. = 0.4*0.6 = 0.24
P(NW,W)  Probability of not winning then winning. 0.6*0.4=0.24
Then we add the two probability together since we want the probability he wins only one game, and there are only two paths, if you like (outcomes) that can lead to Neil only winning one game.
So, it's 0.24+0.24 = 0.48
2A) So it wants the probability that neither at school, the probability that Anna isn't at school is 0.4, and the probability that Bill isn't at school is 0.3
It's conditional probability, since the probability of Bill going to school (or not going to school) depends on what Anna does, therefore we multiply.
0.4*0.3 = 0.12  Which is the final answer since that is the only path (outcome) in which both Anna and bill DON'T go to school.
2B) At least one of them goes to school.
So it wants to know the probability that what the probability of Anna going to school when Bill bill does not, and when Anna doesn't go to school, but Bill does.
(2 possible outcomes [paths])
P(A,DA)  Probability that Anna attends but Bill does not attend. = 0.6*0.3 = 0.18
P(DA,A)  Probability that Anna does not attend but Bill does attend = 0.4*0.7=0.28
Now we add the two possible outcomes together to get 0.18+0.28 = 0.46
(Which you got correct!) 
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 25062016 12:06
(Original post by 34908seikj)
For A, you multiply not ADD because the probability of winning or not winning the second game depends on the outcome of the first game.
Instead of 0.4+0.4, it would be 0.4*0.4 = 0.16
Which you probably just wrote addition by mistake, given that you got the right answer!
Part B, You need, the probability he will win in game one * by the probability that he does not win game two.
Then you also need to probability he does not win a game * the probability that he wins a game. Then just add the two together to get your answer
P(W,NW)  Probability of winning then not winning. = 0.4*0.6 = 0.24
P(NW,W)  Probability of not winning then winning. 0.6*0.4=0.24
Then we add the two probability together since we want the probability he wins only one game, and there are only two paths, if you like (outcomes) that can lead to Neil only winning one game.
So, it's 0.24+0.24 = 0.48
2A) So it wants the probability that neither at school, the probability that Anna isn't at school is 0.4, and the probability that Bill isn't at school is 0.3
It's conditional probability, since the probability of Bill going to school (or not going to school) depends on what Anna does, therefore we multiply.
0.4*0.3 = 0.12  Which is the final answer since that is the only path (outcome) in which both Anna and bill DON'T go to school.
2B) At least one of them goes to school.
So it wants to know the probability that what the probability of Anna going to school when Bill bill does not, and when Anna doesn't go to school, but Bill does.
(2 possible outcomes [paths])
P(A,DA)  Probability that Anna attends but Bill does not attend. = 0.6*0.3 = 0.18
P(DA,A)  Probability that Anna does not attend but Bill does attend = 0.4*0.7=0.28
Now we add the two possible outcomes together to get 0.18+0.28 = 0.46
(Which you got correct!)
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 25062016 12:11
(Original post by 34908seikj)
For A, you multiply not ADD because the probability of winning or not winning the second game depends on the outcome of the first game.
Instead of 0.4+0.4, it would be 0.4*0.4 = 0.16
Which you probably just wrote addition by mistake, given that you got the right answer!
Part B, You need, the probability he will win in game one * by the probability that he does not win game two.
Then you also need to probability he does not win a game * the probability that he wins a game. Then just add the two together to get your answer
P(W,NW)  Probability of winning then not winning. = 0.4*0.6 = 0.24
P(NW,W)  Probability of not winning then winning. 0.6*0.4=0.24
Then we add the two probability together since we want the probability he wins only one game, and there are only two paths, if you like (outcomes) that can lead to Neil only winning one game.
So, it's 0.24+0.24 = 0.48
2A) So it wants the probability that neither at school, the probability that Anna isn't at school is 0.4, and the probability that Bill isn't at school is 0.3
It's conditional probability, since the probability of Bill going to school (or not going to school) depends on what Anna does, therefore we multiply.
0.4*0.3 = 0.12  Which is the final answer since that is the only path (outcome) in which both Anna and bill DON'T go to school.
2B) At least one of them goes to school.
So it wants to know the probability that what the probability of Anna going to school when Bill bill does not, and when Anna doesn't go to school, but Bill does.
(2 possible outcomes [paths])
P(A,DA)  Probability that Anna attends but Bill does not attend. = 0.6*0.3 = 0.18
P(DA,A)  Probability that Anna does not attend but Bill does attend = 0.4*0.7=0.28
Now we add the two possible outcomes together to get 0.18+0.28 = 0.46
(Which you got correct!)
I want to do it myself but I need a bit of explanation
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 25062016 12:30
(Original post by z_o_e)
So for this...
I want to do it myself but I need a bit of explanation
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So what are the possible combinations of disk one and disk two having the same colour?
Well Disk one can be read, and disk two can be also red, as well as disk one being white and disk two also being white, we can notate this like so:
P(R,R)
P(W,W)
To work out the probability of them both being the same colour, you'll have to work out the probability of both red then both white, adding them together to get your final answer.
B) At least one of the disks was red.
Exactly the same thing as in part A. However Instead of working out the probability for R,R and W,W, you'll be trying to work out R,W and W,R 
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 25062016 13:10
(Original post by 34908seikj)
A) Both disks are the same colour.
So what are the possible combinations of disk one and disk two having the same colour?
Well Disk one can be read, and disk two can be also red, as well as disk one being white and disk two also being white, we can notate this like so:
P(R,R)
P(W,W)
To work out the probability of them both being the same colour, you'll have to work out the probability of both red then both white, adding them together to get your final answer.
B) At least one of the disks was red.
Exactly the same thing as in part A. However Instead of working out the probability for R,R and W,W, you'll be trying to work out R,W and W,R
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 25062016 13:44
It should be 6/9 for red and 3/9 for white because you have 6 disk out of a total of 9 and the same applies to the white disks, 3 white disks out of 9 total.
So, disk one the probability of getting red is 6/9 and for white it is 3/9, which can be simplified to 1/3.
Try and do the questions again with the corrected probabilities, you method was right though.
A) R,R = 6/9 * 6/9 = 36/81
W,W = 3/9 * 3/9 = 9/81
(36+9)/81 = 45/81, Which can be simplified to 5/9.
Try it for part B! 
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 25062016 14:14
(Original post by 34908seikj)
You've messed up on the probabilities. Disk one, you say "6" for R and "3" for white... what does that mean, in terms of probability?
It should be 6/9 for red and 3/9 for white because you have 6 disk out of a total of 9 and the same applies to the white disks, 3 white disks out of 9 total.
So, disk one the probability of getting red is 6/9 and for white it is 3/9, which can be simplified to 1/3.
Try and do the questions again with the corrected probabilities, you method was right though.
A) R,R = 6/9 * 6/9 = 36/81
W,W = 3/9 * 3/9 = 9/81
(36+9)/81 = 45/81, Which can be simplified to 5/9.
Try it for part B!
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 25062016 14:17

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 25062016 15:45
(Original post by 34908seikj)
Which question is this? Why did you work out the probability for getting one red disk, then the probability of getting two white?
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Updated: December 10, 2016
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