Basic Math Challenge - May 2018

[fresh_42]

... and the $\mathbb{F}_8-$basis over $\mathbb{F}_2$ with its multiplication rules.

 

[Zafa Pi]

Problem 6 is still open.

I find #6 ambiguous. If an urn has 3 whites and 1 black what do you think is the probability of drawing 3 that are either all white or all black?

 

[Zafa Pi]

Problem 6 is still open.

I'd be much obliged if someone tackles it

The only way I can make a reasonable problem from the way it was posed is as follows:
The probability of getting all white or all black from urn 2 is prob all white + prob all black.
In that case it is impossible due to Fermat's Theorem and n ≥ 3.

That's what Fermat told me.

 

[StoneTemplePython]

The only way I can make a reasonable problem from the way it was posed is as follows:
The probability of getting all white or all black from urn 2 is prob all white + prob all black.
In that case it is impossible due to Fermat's Theorem and n ≥ 3.

That's what Fermat told me.

Can you expand on this a bit for others' benefit?

It's a mildly famous problem designed by E. C. Molina, to cast Fermat's Last Theorem as a probability problem.

 

[mfb]

There are some trivial solutions for (6).

If an urn has 3 whites and 1 black what do you think is the probability of drawing 3 that are either all white or all black?

1/4³ + (3/4)³. We draw with replacement.

In that case it is impossible due to Fermat's Theorem and n ≥ 3.

Huh?

 

[Zafa Pi]

Huh?

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.

 

[mfb]

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]

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.

 

[mfb]

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]

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

 

[fresh_42]

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.