Red, blue and green

Probability Level pending

A bag contains 100 red balls, 10 green balls, and some blue balls.

How many blue balls would maximize the probability that you could pick 3 balls without replacement and they would all end up being different colors (i.e. one red, one blue, and one green)?

If there are multiple answers, please provide the answer as the sum of all such answers. That is, if $a$ and $b$ both maximize the probability, the answer would be $a+b$ .

Assume that there are no other colors in the bag.

The answer is 109.

1 solution

Geoff Pilling
Sep 3, 2016

The formula that represents this probability is:

$\large P(n) = \frac{100 \times 10 \times n}{\binom{100+10+n}{3}}$

This is maximized for $n = 54$ and $55$ .

$54+55 = \boxed{109}$

Is this solvable without Excel?

Pi Han Goh - 4 years, 9 months ago

$P'(n) = 0$ and solve for $n$ . If it's an integer you are done. If it isn't try the ceiling and floor of $n$ . If they match, you add them up, otherwise take the smallest of the two.

Geoff Pilling - 4 years, 9 months ago

you need to show that P''(n) < 0 as well, but yeah. I thought you got a better approach hiding in your sleeves.

Pi Han Goh - 4 years, 9 months ago

@Pi Han Goh – Ah... good point!

Geoff Pilling - 4 years, 9 months ago

Red, blue and green

The answer is 109.

1 solution

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