Melody

Please note that I have edited my answer and it is open to examination  

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

P(2 twos and 2 ones and the other two different ) +  P(2 sixes and 2 ones and 2 of some other number)

\( =\frac{6!}{2!2!}*(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{4}{6}*\frac{3}{6}) + \frac{6!}{2!2!2!}*(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{4}{6}*\frac{1}{6})\\ =180*\frac{12}{6^6} + 90*\frac{4}{6^6}\\ =\frac{2160}{6^6} + \frac{360}{6^6}\\ =\frac{2520}{46656}\\ =\frac{35}{648} \\\approx 0.054\)

I do not think this is correct, I think 2160 should be half the size at 1080.

BUT I do not know what I did wrong. Probablility is like that ... it's sneaky!!

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Here is another way to look at it 

You have six squares an each square represents a number rolled.

There will be 6^6 = 46656  possible outcomes but I have to work out how many of those are favourable.

First there are 6C2 = 15 places where the 1s can go.

Now there are four spots left so

There are  4C2= 6 places where the 2s can go.

So far that is   15*6=90

Now there are 2 spots left.

If the the last two numbers are different then you can chose from 3,4,5, or 6

So there will be 4 choices for the next place and 3 fro the last place so that is    90*4*3 = 1080

If the the last two numbers are the same then you can chose from 3,4,5, or 6

So there will be 4 choices for the next place and 1 for the last place so that is    90*4*1 =360

So that is 1080+360 = 1440

So the probablility is \(\frac{1440}{6^6}=\frac{1440}{46656}=\frac{5}{162}\)

This is why I think that the second answer is correct. I counted the possibilities.

5w^-2 y^6 * 2w^5 y

\(5w^{-2} y^6 \times 2w^5 y\\5w^{-2} y^6 \times 2w^5 y^1\\ =5\times 2w^{-2+5} y^{6+1}\\ =10w^{3} y^{7}\\\)

Thanks very much Heureka :)

What a cute puppy !!

It is good that you are back :))

Hi Hayley   

Well it doesn't make sense to me, it says that division gives a polynomial g(x)

"Let f(x) = 3(x^4+x^3+x^2+1)/(x^2+x-2). Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of 0 as x approaches positive infinity."

Was this the intended question?

Let f(x) = 3(x^4+x^3+x^2+1)/(x^2+x-2).

Find a polynomial g(x) such that f(x) + g(x) has a horizontal asymptote of 0 as x approaches positive infinity.

But Alan

 3(x^4+x^3+x^2+1)/(x^2+x-2) 

does not divide to give a polynomial so how can you say that this question makes sense.  ?