ChatterBank66 mins ago
G C S E And Even More Sweets
8 Answers
In a bag there are some red sweets and some green sweets.
The ratio of red sweets to green sweets is 2:3
Two sweets are taken at random and eaten.
The probability they're both green is 35%
How many sweets were in the bag before any sweets were removed.
The ratio of red sweets to green sweets is 2:3
Two sweets are taken at random and eaten.
The probability they're both green is 35%
How many sweets were in the bag before any sweets were removed.
Answers
1 to 8 of 8
In some ways I'm the complete opposite of prudie -- I find the algebra in these problems less stimulating than the set-up, since by that point you're following rules to such an extent that, in effect, the algebra is just doing itself and you're along for the ride. In that sense, this is the same problem as the previous series of sweets questions, because the...
13:50 Mon 17th Apr 2023
In some ways I'm the complete opposite of prudie -- I find the algebra in these problems less stimulating than the set-up, since by that point you're following rules to such an extent that, in effect, the algebra is just doing itself and you're along for the ride.
In that sense, this is the same problem as the previous series of sweets questions, because the set-up is the same.
How do I read these questions? When I'm looking at it, I confess, my eyes glazed over most of the details, and I was just looking at various key words. In rough order, these were:
1. "there are some red and green sweets", where the only part I was really caring about was that there are TWO types of thing (again, I capitalise for emphasis not to shout).
2. Then I see "the ratio of..." and more or less stop reading, because I understand that this will be useful information later to relate the number of each type of thing to the total. I'll think about how to do that later.
3. After that, I read "two ... taken at random and eaten", so I recognise this is the exact same set-up as before. The "eaten" here is important because that means that the number of total sweets changes, so this is a "without replacement probability" problem.
4. Then I see "they're both [one type]", and again I stop, I don't care about the number.
So my breakdown of the problem is "two types... in some ratio... taken at random and [not replaced] ... both one type ... solve."
It's obviously not the *only* way you can solve this problem, but what I am trying to emphasise is the key information needed to get into this.
My advice to your grandson, if he's floundering in trying to get into the algebra, is to try and seek these key words. So the aim is to relate to what you've seen before, but *also* to understand how to use what you've seen before in a new way. So:
1. You have R red sweets and G green sweets.
2. The probability of eating two of the same sweet in a row is
(G/N)*(G-1)/(N-1)
if green (or (R/N)*(R-1)/(N-1) if red), and we are told that
(G/N)*((G-1)/(N-1)) = 0.35 = 7/20
Again, because I obsess about generality, I barely care about the actual probability, but that is a matter of taste, and if it helps to go concrete then by all means ignore my choice here.
I think it helps, though, to rearrange at this point and forget what else we know about red/green/total information. You get:
20G(G-1) = 7N(N-1) (1)
and a key point I'm trying to emphasise is that you can get to this point for literally all "two sweets" problems with no real "thought". The only question you now have to ask is "what is G in terms of N" (or vice versa).
Using the information given:
R + G = N
R:G = 2:3 => R = 2G/3
=> (2/3) G + G = (5/3) G = N
and now substitute in to (1). The algebra from there is more of the same, another quadratic equation in G, with some whole-number solution, that will be somewhere in the teens probably, so that N is in the region of 20-something. In this particular case, it's probably helpful to note that if R:G = 2:3 and R and G are both whole numbers, then N is a multiple of 5, ie your candidate solutions are N=15, N= 20, N=25, N=30 etc., although I wouldn't try using this method in the actual exam, more as a sanity check.
I leave it to Prudie, or anyone else who's interested, to do the algebra and confirm the solution (N=25), but as I understand from your last question, the algebra itself isn't the issue, so I don't think it's helpful for me to provide it. That's why I've burbled on and on at length about my approach to setting up the problem. It boils down in the end to seeing enough key words to write down the formula (1) above, and then think about how to relate Y and N for the specific problem.
In that sense, this is the same problem as the previous series of sweets questions, because the set-up is the same.
How do I read these questions? When I'm looking at it, I confess, my eyes glazed over most of the details, and I was just looking at various key words. In rough order, these were:
1. "there are some red and green sweets", where the only part I was really caring about was that there are TWO types of thing (again, I capitalise for emphasis not to shout).
2. Then I see "the ratio of..." and more or less stop reading, because I understand that this will be useful information later to relate the number of each type of thing to the total. I'll think about how to do that later.
3. After that, I read "two ... taken at random and eaten", so I recognise this is the exact same set-up as before. The "eaten" here is important because that means that the number of total sweets changes, so this is a "without replacement probability" problem.
4. Then I see "they're both [one type]", and again I stop, I don't care about the number.
So my breakdown of the problem is "two types... in some ratio... taken at random and [not replaced] ... both one type ... solve."
It's obviously not the *only* way you can solve this problem, but what I am trying to emphasise is the key information needed to get into this.
My advice to your grandson, if he's floundering in trying to get into the algebra, is to try and seek these key words. So the aim is to relate to what you've seen before, but *also* to understand how to use what you've seen before in a new way. So:
1. You have R red sweets and G green sweets.
2. The probability of eating two of the same sweet in a row is
(G/N)*(G-1)/(N-1)
if green (or (R/N)*(R-1)/(N-1) if red), and we are told that
(G/N)*((G-1)/(N-1)) = 0.35 = 7/20
Again, because I obsess about generality, I barely care about the actual probability, but that is a matter of taste, and if it helps to go concrete then by all means ignore my choice here.
I think it helps, though, to rearrange at this point and forget what else we know about red/green/total information. You get:
20G(G-1) = 7N(N-1) (1)
and a key point I'm trying to emphasise is that you can get to this point for literally all "two sweets" problems with no real "thought". The only question you now have to ask is "what is G in terms of N" (or vice versa).
Using the information given:
R + G = N
R:G = 2:3 => R = 2G/3
=> (2/3) G + G = (5/3) G = N
and now substitute in to (1). The algebra from there is more of the same, another quadratic equation in G, with some whole-number solution, that will be somewhere in the teens probably, so that N is in the region of 20-something. In this particular case, it's probably helpful to note that if R:G = 2:3 and R and G are both whole numbers, then N is a multiple of 5, ie your candidate solutions are N=15, N= 20, N=25, N=30 etc., although I wouldn't try using this method in the actual exam, more as a sanity check.
I leave it to Prudie, or anyone else who's interested, to do the algebra and confirm the solution (N=25), but as I understand from your last question, the algebra itself isn't the issue, so I don't think it's helpful for me to provide it. That's why I've burbled on and on at length about my approach to setting up the problem. It boils down in the end to seeing enough key words to write down the formula (1) above, and then think about how to relate Y and N for the specific problem.
It, hopefully, goes without saying that, although I'm contrasting my approach with Prudie's, I'm certainly not intending to mark out one as better than the other. We all find ways that work, or not, to solve problems. My concern is that seeing concrete examples solved over and other, that differ only by the tiny variations, doesn't help your grandson appreciate how to overcome the obvious problem that there will be yet another variation on the theme in the exam proper. I think in sequence, we've seen:
1) the number of the eaten type of sweet is given;
2) the number of the *uneaten* type of sweet is given;
3) the *ratio* of the two types of sweets is given.
A further problem might give, say:
i) the *difference* between the two types of sweet;
ii) the total N, and you might be asked to work out how many of the (un)eaten type of sweet there are;
iii) a different sequence of eaten sweets, eg yellow then green rather than both one or the other;
iv) and so on.
If each example is seen as a separate problem, then these slight variations in what you're asked for could be enough to throw you. But they're all the same in the end. you have two equations:
(a) R+G = N ;
(b) G(G-1)/N(N-1) = p (given probability, 35% in this case) ;
and the third piece of information: R = given, or G = given, or N = given, or R/G = given, or R-G equals given, or whatever.
1) the number of the eaten type of sweet is given;
2) the number of the *uneaten* type of sweet is given;
3) the *ratio* of the two types of sweets is given.
A further problem might give, say:
i) the *difference* between the two types of sweet;
ii) the total N, and you might be asked to work out how many of the (un)eaten type of sweet there are;
iii) a different sequence of eaten sweets, eg yellow then green rather than both one or the other;
iv) and so on.
If each example is seen as a separate problem, then these slight variations in what you're asked for could be enough to throw you. But they're all the same in the end. you have two equations:
(a) R+G = N ;
(b) G(G-1)/N(N-1) = p (given probability, 35% in this case) ;
and the third piece of information: R = given, or G = given, or N = given, or R/G = given, or R-G equals given, or whatever.
If these things were easy to do then KB's grandson wouldn't need help, and if they were easy to explain then there wouldn't be a shortage of maths teachers.
I think the length is justified because I'm trying to repeat myself a few times, in the hope that reading more or less the same thing a few times over helps it sink in a bit more. Also, what's missing from the process is interaction. Who know how helpful any of my answers are, to the person who matters? Maybe not at all, and if so I apologise. But the extra tip I'd add is that, sometimes, answers that long aren't meant to be read all at once, or just one time. Maybe revisiting it a few times, or breaking it into chunks, will help it be absorbed better.
I think the length is justified because I'm trying to repeat myself a few times, in the hope that reading more or less the same thing a few times over helps it sink in a bit more. Also, what's missing from the process is interaction. Who know how helpful any of my answers are, to the person who matters? Maybe not at all, and if so I apologise. But the extra tip I'd add is that, sometimes, answers that long aren't meant to be read all at once, or just one time. Maybe revisiting it a few times, or breaking it into chunks, will help it be absorbed better.
Thankyou Clare!! Your posts are always well received and 'we' are very grateful!!
Zebu on occasion can be a little acerbic, I think it's more impatience on his part, but he means no harm!!
I'd only been on here a few days and he sent me off with a flea in my ear. See last post of the thread (link) below:
https:/
It's just his way LOL :-))
A big thank you to all contributors. Keep the answers coming!!
Zebu on occasion can be a little acerbic, I think it's more impatience on his part, but he means no harm!!
I'd only been on here a few days and he sent me off with a flea in my ear. See last post of the thread (link) below:
https:/
It's just his way LOL :-))
A big thank you to all contributors. Keep the answers coming!!
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