Maths year 11
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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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I can't decrease for some reason...
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(edited 7 years ago)
Original post by 34908seikj
Don't I divide 2726 by 0.73?
Decrease?
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Original post by z_o_e
dividing by a number less than one would increase the number, multiplying a number less than one would decrease the number!
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!
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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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 ?
Hi,
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.
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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If the decimal is bigger than 1st divide.
If it is smaller than 1 I multiply.
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Yes. But it does depend on the question.
Original post by 34908seikj
Yes. But it does depend on the question.
Heyaax
For this do I multiply or add the values to find the answers.
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Original post by z_o_e
HeyaaxFor this do I multiply or add the values to find the answers.Posted from TSR Mobile
For part A, multiply. For part B, multiply then add.
How'd the exam go?
Original post by 34908seikj
For part A, multiply. For part B, multiply then add.
How'd the exam go?
How'd the exam go?
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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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...
Posted from TSR Mobile
Terribly hard.
Lower bound and upper bound came up.
Locai.
Hard equations...
Posted from TSR Mobile
(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.
Original post by 34908seikj
For part A, multiply. For part B, multiply then add.
How'd the exam go?
How'd the exam go?
Can you check these please?
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Original post by z_o_e
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!)
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!)
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!)
Now I get it!!
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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!)
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!)
So for this...
I want to do it myself but I need a bit of explanation
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Original post by z_o_e
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
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
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
I got this
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Original post by z_o_e
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!
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!
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!
I got this
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Original post by z_o_e
Which question is this? Why did you work out the probability for getting one red disk, then the probability of getting two white?
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?
So I need to work out one red and one white and add them up?
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