Wikipedia:Reference desk/Archives/Mathematics/2019 June 16

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

June 16

Probability calculation

Im a bit confused regarding an equation. I looked at the question "what is the probability of drawing 4 cards of the same colour from a deck of cards".

Using the multiplication principle the easy answer should be 26/52 x 25/51 x 24/50 x 23/49 which should give 5.5%

However everyone else seems to be getting a different answer, can anyone tell me where im going wrong?

91.101.26.175 (talk) 16:38, 16 June 2019 (UTC)

What is the probability that, when drawing 1 card at random, all cards drawn have the same color? --JBL (talk) 16:41, 16 June 2019 (UTC)

The first term 26/52 that the OP quotes is the probability of drawing the first card with a predefined colour. However the question does not predefine a colour. It only requires that cards nos. 2- 3 - 4 have the same colour as whatever no. 1 card turned out to be. Omit that term and the result is 11.04..% . DroneB (talk) 15:04, 17 June 2019 (UTC)

That makes sense but the result other people are getting at is 0,01035 or something close to it which would barely be 1%, so that makes our results even further apart. Some of the people arrived at that number independently others got there by multiplying the chance of getting 4 cards of a specific colour in a row (0,002588) which i dont get where that 4 multiplication factor comes from.

So would it be correct to say that getting 4 cards of a specific colour is 5,5% and getting 4 cards of the same but random colour is 11%? 91.101.26.175 (talk) 21:48, 18 June 2019 (UTC)

For four cards of the same color (i.e. the last three cards are the same color as the first), should be 25/51 x 24/50 x 23/49 = 11.044%. And four cards of a specified color is about 5.52%. Bubba73 ^{You talkin' to me?} 21:56, 18 June 2019 (UTC)

I guess there is confusion between the two colors red and black versus the four suits diamonds (♦), clubs (♣), hearts (♥) and spades (♠). The chance of four cards in the same not predefined suit is 12/51 × 11/50 × 10/49 = 0.01056. I noticed your IP address is Danish and you sometimes use a decimal comma like in Denmark. I'm Danish. The Danish name for suit is kulør which can also mean color (farve in Danish) but has a different meaning in cards. If you saw a Danish text saying kulør then 0,01035 is nearly right and you mistranslated the question. PrimeHunter (talk) 18:21, 23 June 2019 (UTC)

Integral domains without nonzero prime ideals in ZF without choice

In ZF without the axiom of choice, is it relatively consistent that there exist integral domains that are not fields but have the zero ideal as their only prime ideal? If so, then any commutative ring with a quotient isomorphic to such a domain would have a prime ideal that is maximal among prime ideals but not among proper ideals, and the Krull dimension would be ill-behaved without choice. In particular, such a domain would be zero-dimensional without being a field, and "zero-dimensional" would not necessarily mean that every prime ideal is maximal qua proper ideal, but only that no prime ideal is contained in any other prime ideal (though it may still be contained in some other proper non-prime ideal). GeoffreyT2000 (talk) 19:36, 16 June 2019 (UTC)

Cubics Comment

Please could someone talk me through multiplying out a cubic, I don't know how many times stuff needs to be multiplied, and frankly, I'm rather confused about the multiplying process. FYI: I know plenty about how cubics work, just the multiplying it out confuses me. An example being (x+3)(x+6)(x-5). Regards, Willbb234 (talk) 20:31, 16 June 2019 (UTC)

There should be one $x^{3}$ term, three $x^{2}$ terms, three $x$ terms and one constant, leaving 8 terms to be added up. For $(x+3)(x+6)(x-5)$, multiplying it out gets you $x^{3}$, $-5x^{2}+6x^{2}+3x^{2}$, $-30x-15x+18x$ and $-90$, which gives us $x^{3}+4x^{2}-27x-90$. Hope this helps. Iffy★Chat -- 20:45, 16 June 2019 (UTC)

Thank you, that helps a lot. I will remember that there are always 8 terms as a result, so that if I get more or less, I know I have gone wrong. Thanks once again. Willbb234 (talk) 20:58, 16 June 2019 (UTC)

@Willbb234: Any similar question (where you vary the number of factors or the number of terms in each factor) can be handled as follows: at each step, do a single application of the distributive law for one product over a sum. For example, with the example you gave, we could do something like ${\begin{aligned}(x+3)(x+6)(x-5)&=((x+3)\cdot x+(x+3)\cdot 6)(x-5)\\&=(x\cdot x+3\cdot x+x\cdot 6+3\cdot 6)(x-5)\\&=(x^{2}+9x+18)(x-5)\\&=x^{2}\cdot (x-5)+9x\cdot (x-5)+18\cdot (x-5)\\&=x^{2}\cdot x+x^{2}\cdot (-5)+9x\cdot x+9x\cdot (-5)+18\cdot x+18\cdot (-5)\\&=x^{3}-5x^{2}+9x^{2}-45x+18x-90\\&=x^{3}+4x^{2}-27x-90,\end{aligned}}$

where in the first step we distribute the first product over the sum (x + 6), in the second step we distribute two different products over the sum (x + 3), in the third step we group like terms, in the fourth step we distribute over the sum (x^2 + 9x + 18), in the fifth step we distribute three different products over the sum (x - 5) = (x + (-5)), and finally we do the arithmetic and combine like terms. This procedure always works; as you become more comfortable with it, you will find yourself using fewer intermediate steps. --JBL (talk) 23:45, 16 June 2019 (UTC)