Melody

Hello again Ginger,   

It is this statement where your logic fails

"The problem, now, is by subtracting 52/11400, it subtracts both wins.  So, what should the value be?"

Look at this simplified example

the number in A is  10+3=13

The number in B is 15+3=18

The number in both is 3

Then number altogether is  13+18-3 = 28

I had to subtract 3 because it had been added in twice :)

Thanks for the compliment Chris :)

Thanks Tiggsy :)

I am always pleased when I see you here and Ginger and I seemed to need an arbitrator here.  

added

I've just looked at your answer properly and it is very easy to follow. (Easier than mine)

Thanks :)

Sorry guest but Hectictar is correct   :)

By convention

\(\sqrt{y^2}=+y\)

that is true

BUT this is different becasuse the square root sign is not in the question. So

If

\(y^2=4\\ then\\ y=\pm\sqrt4=\pm2\)

Look here

0.25 is  25/100 and that simplies to 1/4

If you have 25c then you have  $0.25 and that is 1/4 of a dollar becasue  25c+25c+25c+25c = 100c = $1

1.0 is just 1

 1/4=0.25

Thanks Ginger, I'm going to try answering independently of you and  see if we get the same answer :))

To answer Melody's question, you pick three numbers from 1 to 20. Once you pick a number from there, it can not be chosen again. You also pick a number from 21 to 30. You pick these numbers at random. ( I never played lottery before, so I'm doing my best.) Again, three numbers from 1 to 20 and one number from 21 to 30 is chosen. These numbers are the winning numbers. If it happens that you match at least two of the numbers from 1 to 20 with the winning numbers, or you match the Superball number, you win a prize. ( The order doesn't matter)

ok, I am going to pretend that the winning numbers are chosen first.  So there are 3 low winning numbers but you only need two of them OR one high winning number and that is the one you need. This will make no difference to the outcome.

Plus there are 2 independent events happening here.

The chance of you getting the high winning number is 1 in 10 that is the easy bit

The number of ways that 3 numbers can be chosen from 20 is 20C3 = 1140   just like Ginger said.

The number of ways that I can get all three numbers is 1

I can get 3 pairs of 2 numbers , just as Ginger said.

The number of ways i can get any one of these pairs and not the third winning number is  17

17*3= 51

So I think there are 51 ways of chosing 2 winning small numbers and not not getting the third.

So that makes a total of 52 ways of getting a winning set of smal numbers.

So I think that the chance of getting 2 or 3 small numbers + getting the big one is

\(\frac{52}{1140}\times\frac{1}{10}=\frac{52}{11400}\)

Which means that the chance of winning is  

\(\frac{52}{1140}+\frac{1}{10}-\frac{52}{11400}\\ =\frac{520}{11400}+\frac{1140}{11400}-\frac{52}{11400}\\ =\frac{520+1140-52}{11400}\\ =\frac{1608}{11400}\\ =\frac{67}{475}\\ \approx 0.141 \)

Ginger's answer and mine are only slightly different. I think it is just rounding error.

If so mine is closer because i didn't round anything until the very end.  

3x(-2)

depends what you mean...

\(3\times (-2)=-6\\ 3x\times (-2)=-6x\\ 3x^{-2}=\frac{3}{x^2}\\ (3x)^{-2}=\frac{1}{9x^2} \)

Hi Ginger :)

Thank you.

I have not tried to answer yet, but the rules seem to be clear now :)

Hi Guest,

Thanks for your answer,

Why don't you start learning LaTex?

It is not that hard, there is always people here that will help you get started.

You can start just by using it for fractions and very basic things and progress from there.

To write   \(5\frac{2}{3}\)

You just open the LaTex tab in the ribbon. Delete the formula that is already there and type in

5\frac{2}{3}          and press  ok :)