Maths help

M

Muffin

New member
Joined
Nov 22, 2023
Messages
1
Q1: Harry has 8 red,4 blue and 4 yellow marbles in his pocket. What is the smallest number of marbles he must take out from his pocket to be sure that he gets at least one of each colour?

Q2: A box contains 100 balls. 28 red,20 green, 12 yellow,20 blue, 10 black and 10 purple. How many balls must you draw from the box to be sure that you will get 15 balls of the same colour?

Can anyone help and show me how to work it out please? Thank you
 
Muffin said:
Q1: Harry has 8 red,4 blue and 4 yellow marbles in his pocket. What is the smallest number of marbles he must take out from his pocket to be sure that he gets at least one of each colour?
Can anyone help and show me how to work it out please? Thank you
Click to expand...
Think about it, @Muffin, you want to get at least one of each colour but the first marbles you draw out might all be the same colour and, if they're all red then you could draw out eight marbles all the same (red) colour!

The ninth marble would then have to be a different colour, blue or yellow, (because there are no red ones left) but what if the tenth marble was the same colour as the ninth?

Let's say the ninth and tenth ones drawn were both blue, how many more would you have to draw to be (absolutely) sure that you finally got a yellow one?

Muffin said:
Q2: A box contains 100 balls. 28 red,20 green, 12 yellow,20 blue, 10 black and 10 purple. How many balls must you draw from the box to be sure that you will get 15 balls of the same colour?
Can anyone help and show me how to work it out please? Thank you
Click to expand...
Again, this requires some thought. Keep in mind that you are trying get 15 balls all of the same colour.
(So they cannot be yellow, black or purple. Why not?)

There are 6 different colours of the balls in the box, so, worst case, the first six you draw out might all be different colours! Then, the next six might all be different colours too; that means that you have drawn 12 balls but only have 2 of each colour!

That could keep happening (drawing 6 different colours each time) right up until you have drawn 60 balls out.
Can you see why?
How many black ones were in there? And how many purple ones?

Once you had drawn 60 in this unlucky fashion (each 6 drawn were all different colours), how many of each colour would you now have?

What is now left in the box? (How many of each colour are left in the box?)

How many could you now draw that were all different colours?
And how long could you keep doing that?

(Hint: There are only two yellow ones left in there.)

Can you now finish the process yourself?

Please come back and tell us what answer you got for each question (or let us know if you're still stuck).

We will then tell you if your answers are correct but if you really can't finish either question then please jot down on a piece of paper the working that you have tried to get to an answer and upload a picture of it so we can see where you are going wrong.

Hope that helps. 😊
 
You want to consider the worst case scenario. As The Highlander said, what if your 1st picks were all red and then your next picks were all blue. What would the next color have to be? Can you do any worse than that?
 
Steven G said:
You want to consider the worst case scenario. As The Highlander said, what if your 1st picks were all red and then your next picks were all blue. What would the next color have to be? Can you do any worse than that?
Click to expand...
Surely, the question should be: "Can you do any better than that?" (With certainty.) 🤣
 
Steven G said:
Can you do any worse than that?
Click to expand...
The Highlander said:
Surely, the question should be: "Can you do any better than that?" (With certainty.) 🤣
Click to expand...

I agree with @Steven G; we are looking for the worst case. Why? Because we want to be sure we get at least one of each color, no matter what happens. I like to imagine that we aren't picking balls at random (and just being unlucky), but they are being given to us by an enemy who wants to frustrate our goal as long as possible. "Can you do any worse?" means "Can you be any more unlucky?"; or, "Can the enemy, who is trying to produce the worst case for you, do better?" Once we know what is the worst he can do, we have solved the problem.

The enemy might give us all the reds and all the blues, and we still won't have one of each color. But the next one he gives us, he loses: We will have reached our goal.

But if we'd first started by supposing that he gave us all the blues and all the yellows, then we'd have to ask that question: Could the enemy have made it take even longer to reach our goal? Then we'd realize that the enemy has a better strategy (namely, giving use the reds and blues first).

I think that's what "worse" means: What is the worst the enemy can throw at us?
 
khansaheb said:
Steven's Demon ?!!!
Click to expand...
I've often called it a gremlin. In particular problems, it might be a sock-drawer demon, an urn imp, or here, perhaps, a pocket pixie. For me, at least, personalizing the dark side of probability makes this sort of problem more interesting.
 
What a kerfuffle over a small joke I was making about semantics! 🤷‍♂️

The OP whom I suspect is just a child certainly doesn't need any of the further 'clarification' that's been provided after my response at Post #2; in fact, it's only likely to confuse or dumbfound them further if they read it!

I believe I specifically stated that we are looking for the "worst case" scenarios to solve these problems and what Steven described in his post was that worst case: "As The Highlander said, what if your 1st picks were all red and then your next picks were all blue. What would the next color have to be?" so you could not do any "worse" than that as any subsequent marble withdrawn would immediately meet the criterion of the question (as it would have to be yellow and you would then have at least one of each colour)!

My point (which, as I say, was purely semantic and only intended as a gentle reproach) was that, if anything, once having reached the point of 8 red & 4 blue marbles withdrawn and about to pick a final one to meet the stated objective, what one should be considering (if anything) is could you do any "better" than that? In other words would it be possible to withdraw fewer than 13 marbles and (with certainty) still meet the stated objective (ie: a "better" result)? Of course, the logic I already presented made it clear (IMNSHO) that it was not possible to do so!

I do hope the OP only reads the first reply and doesn't get bemused by anything that (redundantly) follows it. 🤔
 
When I wrote 'can you do any worse than that', I was asking a serious question. Just because the next ball had to be yellow doesn't mean (in theory) that this was the worst case scenario. If one first picked all the blue and yellow balls, then red must be next. However, this is not the worst case scenario. As I usually do, I just want the OP to think about the problem and specifically think if there was a scenario worse than the one I hinted towards.
 
You must log in or register to reply here.
Top