Challenge Basic Math Challenge - May 2018

• Featured

Zafa Pi

Let N be the number of balls in each urn, W the number of white balls in urn 1, b and w the number of black and white balls in urn 2, n > 2 the number drawn from each urn. With my interpretation in #73 of the problem we get:
(W/b+w)n = (b/b+w)n + (w/b+w)n.
Multiplying by (b+w)n we have Wn = bn + wn which is impossible for n > 2 by Fermat's (Wiles') Theorem.

• StoneTemplePython

mfb

Mentor
Ah. I interpreted "the probability that all white balls are drawn from the first urn" as probability that every white ball in the urn is drawn at least once.

Zafa Pi

Ah. I interpreted "the probability that all white balls are drawn from the first urn" as probability that every white ball in the urn is drawn at least once.
I think that the way the problem was formulated is problematic. I attempted to point that out to @StoneTemplePython in post #72, but received no response.
For example, If an urn contains 3 whites and one black, then what is the probability of drawing (with replacement) three balls such that they are either all white or all black? I think the only reasonable answer is ¾3, which = the probability of drawing three white balls. Thus urn 1 and urn 2 are the same is a solution.

Perhaps it should be said that the urns contain at least 3 (or n) of each color.

StoneTemplePython

Gold Member
I think that the way the problem was formulated is problematic. I attempted to point that out to @StoneTemplePython in post #72, but received no response.
For example, If an urn contains 3 whites and one black, then what is the probability of drawing (with replacement) three balls such that they are either all white or all black? I think the only reasonable answer is ¾3, which = the probability of drawing three white balls. Thus urn 1 and urn 2 are the same is a solution.

Perhaps it should be said that the urns contain at least 3 (or n) of each color.
This problem and its wording, setting aside one or two nits, is verbatim from Mosteller's Fifty Challenging Problems in Probability. It is the last problem in the book, of course numbered 50 56. (There are some bonus problems.)

People are free to ask questions to clarify, though posts #72 and #73, $\lt 3 \text{ hours }$ apart, both arrived while I was offline.

- - -
My read: Post #72 did not directly seek to clarify the wording of the question but instead said "what do you think...".

On the other hand, Post #73 was direct and had the answer. I responded to the most recent, accurate and direct post, #73.

- - -
n.b.

I also liked this:

That's what Fermat told me.

Last edited:
• fresh_42

mfb

Mentor
Post 73 doesn't address the interpretation question for urn 1. As you marked it as solved I guess the different wording for urn 1 doesn't mean anything. "It is impossible" is a strange solution to "Find the number of [...]".

With the other interpretation I could rule out 1 and 2 white balls in urn 1 for 3 and 4 drawings, but the problem gets much more difficult.

Greg Bernhardt

Just a couple more sub questions to be answered. Incredible works everyone!

fresh_42

Mentor
2018 Award
First day open to all. Open: 1.b.) $\mathbb{F}_2-$ basis of $\mathbb{F}_8$ and its multiplication resp. addition structure.

fresh_42

Mentor
2018 Award
Solution to 1.b)
From there determine a basis of $\mathbb{F}_8$ over $\mathbb{F}_2$ and write down its multiplication and addition laws.
This is actually the easiest part of all. Let's say we have the minimal polynomial $x^3+x+1=0\,$, and we assume a root $\xi \,$, i.e. an element which satisfies $\xi^3+\xi +1 =0$. Then
$$x^3+x+1 = (x+\xi)(x+\xi^2)(x+\xi+\xi^2)$$
and $\{1,\xi,\xi^2\}$ is a $\mathbb{F}_2$ basis of $\mathbb{F}_8$. The elements are thus
$$\{0,1,\xi,\xi+1 = \xi^3,\xi^2,\xi^2+1=\xi^6,\xi^2+\xi=\xi^4,\xi^2+\xi+1=\xi^5\}$$
which defines the multiplicative group generated by $\xi$ as well as the addition table.

"Basic Math Challenge - May 2018"

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