smallest number of balls that must be drawn to ensure that at least 50 balls of one color are selected

ShanQ

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Original question:
A box contains 400 balls, each of which is blue, red, green, yellow or orange.
The ratio of blue to red to green balls is 1 : 4 : 2. The ratio of green to yellow to orange balls is 1 : 3 : 6.
What is the smallest number of balls that must be drawn to ensure that at least 50 balls of one color are selected?

My workings:
Total # of balls = 400
blue : red : green : yellow : orange = 1:4:2:6:12
1+4+2+6+12=25
Therefore, # of balls of each color:
blue: 1/25*400= 16
red: 4/25*400=64
green: 32
yellow: 96
orange: 192

I guess I would do the following calculations:
In order to choose at least 50 balls of one color, it is only possible with {red, yellow, orange}.

Case 1: red
400C1 + 400C2 + … + 400Cx >= 64C50

Case 2: yellow
400C1 + 400C2 + … + 400Cy >= 64C50

Case 3: orange
400C1 + 400C2 + … + 400Cz >= 64C50

Choose a minimum value among {x, y, z}

But we should choose the ball out of 400 randomly.
I don't know how to continue …

Thank you so much for your help.
 
If I interpret the wording of the problem literally, we clearly must must draw AT LEAST 50 balls to ensure fifty balls of the same color.

But is that what the problem is really asking for? Did you quote the problem exactly?
 
Thank you JeffM. I think that wording should be ok. The question is directly from a math textbook from Cambridge.
It says “at least 50 balls of one color”.
If you draw 50 balls from the 400, you cannot guarantee at least 50 balls of the same color.
Now, I am thinking of using pigeonhole principle for this problem. But I haven’t found the solution yet. Thank you for your help.
 
What is the smallest number of balls that must be drawn to ensure that at least 50 balls of one color are selected?
This isn't about randomness; it's about certainty. You want to be sure to pick 50 of the same color.

Suppose a gremlin in the box controlled which ball you drew, and wanted to keep you from reaching the goal. What might he do? When would he be unable to stop you?
 
I still think the wording does not reflect what you perceive the intent to be.

What is the smallest number of balls you can draw such that the probability of at least fifty being of the same color is 1?

That is how Dr. Peterson tries to make the problem non-trivial.

And I agree with his method though I might restate it to think first about the largest number of draws that is consistent with having fewer than fifty of the same color.
 
To be guaranteed 50 of one color you have to pick:
How many blue balls?
How many red balls?
How many green balls?
How many yellow balls?
How many orange balls?
 
The textbook answer is 196. I guess the drawing is done randomly without knowing the color beforehand.
 
The textbook answer is 196. I guess the drawing is done randomly without knowing the color beforehand.
Yes, the balls are drawn randomly; but you are not asked for a probability, but how you can make the outcome certain: that you can be sure that at least 50 will have the same color. The question is about the worst case, not the typical. And 196 is correct.

To repeat the problem:

A box contains 400 balls, each of which is blue, red, green, yellow or orange.​
The ratio of blue to red to green balls is 1 : 4 : 2. The ratio of green to yellow to orange balls is 1 : 3 : 6.​
What is the smallest number of balls that must be drawn to ensure that at least 50 balls of one color are selected?​

You found that there are 16 blue, 64 red, 32 green, 96 yellow, and 192 orange balls.

Suppose, for example, that you draw 100 balls. They MIGHT all be orange; but they could be 16 blue, 23 red, 25 green, 18 yellow, and 18 orange balls -- not 50 of any color. If you drew all 400, then of course there would be more than 50 each of red, yellow, and orange. Somewhere in the middle will be your answer. How can you make sure that a random drawing of N balls will include at least 50 of some color?

Have you been introduced to the Pigeonhole Principle? You can either use that explicitly, or do as I suggested, which I find more intuitive.
 
Thank you Dr Peterson. I knew the Pigeonhole Principle.
This is what I thought below:
I mark 5 boxes: blue, red, green, yellow, and orange colors.
Then I start to draw balls out of the 400 randomly.
I sort the balls according to the color and put each ball into the corresponding colored box.

The max number of blue box can contain = 16 balls
The max number of green box can contain = 32 balls.
The rest 3 boxes: number of balls can be greater >50 balls.

When I draw 16+32+50*3 =198 balls, the blue, green box may be full; if the random balls are not blue or green, they will contribute more to the remaining colors; one of the 3 major colors will be at least 50. Therefore, 198 balls can guarantee that at least 50 balls of one color are selected.

I don't know how to work out 196 balls as the textbook answer.
 
Thank you Dr Peterson. I knew the Pigeonhole Principle.
This is what I thought below:
I mark 5 boxes: blue, red, green, yellow, and orange colors.
Then I start to draw balls out of the 400 randomly.
I sort the balls according to the color and put each ball into the corresponding colored box.

The max number of blue box can contain = 16 balls
The max number of green box can contain = 32 balls.
The rest 3 boxes: number of balls can be greater >50 balls.

When I draw 16+32+50*3 =198 balls, the blue, green box may be full; if the random balls are not blue or green, they will contribute more to the remaining colors; one of the 3 major colors will be at least 50. Therefore, 198 balls can guarantee that at least 50 balls of one color are selected.

I don't know how to work out 196 balls as the textbook answer.
You're very close to the right idea; but you have too many 50's there. You already have 50 each of three different colors, so you're beyond the goal.

If I were handing you the balls as you choose them, and trying not to let you have 50 of any color, I wouldn't give you 50 each of red, yellow, and orange. How many would I give you?

Then, the next ball allows you to "win".
 
When I draw 16+32+50*3 =198
When you have drawn 16+32+49*3 (=195) balls - you do not yet have 50 balls of any color. Then you draw the next ball (196) - you have 50 balls of some color.
 
Thank you Subhotosh Khan. Amazing explanation!
Many thanks to Dr Peterson and others!.
 
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