Random variable problem

[Abd-al-Karim]

I cannot get my head around the following problem:

 

An irregular six face die is thrown and the probability that in twenty throws it will give five even numbers is twice the probability that it will give five odd numbers, i.e.:

 

P(5 even numbers is 20 throws) =2P(5 odd numbers in 20 throws)

=2P(15 even numbers in 20 throws).

 

How many times in 10000 sets of 10 throws would you expect it to give no even number? [Hint ]

 

I do not fully understand it I think they are looking for a rnadom variable with a binomial distribution. Also I am A bit condused by P(5 even numbers is 20 throws) is that is maybe a typo and its supposed to be "in"? Also I don't see how that could be equal to =2P(15 even numbers in 20 throws)

 

Any help would be appreciated. Thanks in advance.

[Aeternus]

Ok heres what i get (but im probably wrong)

 

20C5 * (x)^5 * (1-x)^5 = 2 * 20C5 * (1-x)^5 * (x)^15

 

The combinations cancel out and we are left with -

 

x^5 (1-x)^15 = 2 x^15 (1-x)^5

 

If we take the x's and 1-x 's over to opposite sides we get

 

(1-x)^10 = 2 x^10

 

and then if we take the x^10 over -

 

((1-x)^10)/(x^10) = 2

 

So if we raise each side to (1/10) (hence the 2^(1/10) hint) we get -

 

(1-x)/x = 2^(1/10)

 

(1-x)/x = 1.072

 

1-x = 1.072x

 

1= 2.072x

 

x= 1/2.072 = 0.48263 (Probability of getting an even)

 

Then if you plug this into the equation to check -

 

20C5 x (0.48263)^5 x (0.51737)^15 = 2 x 20C5 x (0.51737)^5 x (0.48263)^15

 

0.0206 = 2 x 0.103 (True)

 

And since 20C5 = 20C15 (as rCp = rC(r-p)), the whole 15 in 20 even throws (x2) thing is also true (as it is basically the same as the 5 in 20 odd).

 

Dont know about the last bit. Perhaps just -

 

10,000 x 10 = 100,000

 

then 100,000 x (1-0.48263) = 51737 ?

 

Although the last bit sounds dodgy and I could easily be wrong.

 

[Edit]

 

Looking at it the whole "in sets of 10" thing seems strange. Does it mean, how many sets of 10 would have no even number??

 

If so, maybe -

 

10C0 x (0.48263)^0 x (0.51737)^10 = 0.51737^10 = 0.001374

 

And then 0.001374 x 10,000 = 13.74 so 13 or 14?

[Abd-al-Karim]

Looking at it the whole "in sets of 10" thing seems strange. Does it mean' date=' how many sets of 10 would have no even number??[/quote']

 

No I am pretty sure they mean that someone throws a dice 10 times 10000 times. I would guess they mean how many sets of 10 have no even number I duppose they want you to work out:

 

for where denotes the number of even numbers. I am not sure about the probability p thats the thing.

 

Thanks for your help so far, I'm just gonna work through your answer now and then I'll get back to you.