Melody

A red and a green dice are rolled. 

What's the probability that the sum of the pips on the top face is 5 or showing number greater than 3 ? 

This is usually the best way to do 2 dice questions - you will always see a pattern emerge.

There are 27 possibilities that have a number greater than 3 They are red.

There are 4 that add to 5 they are highlighted in yellow.

There are 2 that are in both groups.  You do not want to add those twice.

So number of possibilities is      27+4 -2 = 29

So the probability that the sum of the pips on the top face is 5 or at least one of the numbers is greater than 3 is    

                   \(\frac{29}{36}\)

CONGRATULATIONS   ON   YOUR   HALF   MILLENIUM   ANONYMOUS4338       

Ans thanks for giving our guests this great advice too.   

YOU CANNOT APPLY THE QUADRATIC FORMULA TO THAT BECAUSE IT IS NOT AN EQUATION.

IT MUST = 0   !!

Let   M=2000,     G=1.03      R=1.04

Let P be the present value

\(Original\; value    =  P\\ End\; of \;1\; year    =  RP-M\\ End\; of \;2nd\; year =  R(RP-M) - GM\\ End\; of \;2nd\; year =  R^2P- RM - GM\\ End\; of \;2nd\; year = R^2P-M(R+G) \\ End\; of \;3rd\; year =  R(R^2P- M( R+G)) - G^2M\\ End\; of \;3rd\; year =  R^3P- M( R^2+GR) - G^2M\\ End\; of \;3rd\; year =  R^3P- M( R^2+GR+G^2) \\ \)

\(End\; of \;4th\; year =  R(R^3P- M( R^2+GR+G^2))-G^3M \\ End\; of \;4th\; year =  R^4P- M( R^3+GR^2+G^2R)-G^3M \\ End\; of \;4th\; year =  R^4P- M( R^3+GR^2+G^2R+G^3) \\ End\; of \;nth\; year =  R^nP- M( R^{n-1}+R^{n-2}G^1+.......R^1G^{n-2}+G^{n-1}) \\ End\; of \;nth\; year =  R^nP- M*\displaystyle\sum_{k=0}^{n-1}\;R^{n-1-k}G^k\)

NOW for our question the amont at the end of the 20th year =0

\(End\; of \;20th\; year = 1.04^{20}P- 2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k\\ 0 = 1.04^{20}P- 2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k\\ 2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k = 1.04^{20}P\\ 1.04^{20}P=2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k \\ P=\left[2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k \right]\div[ 1.04^{20}]\\\)

Initial value was $35,142.88          

Fri 27/11/15

1) Call for guests to become members.   Thanks Anonymous4338  :)

https://web2.0calc.com/questions/web-2-0-calc

2) How do you know if a number is prime?  Thanks Gino3141 and guest.  :)

https://web2.0calc.com/questions/prime-numbers_2

3) This was in yesterday and I would still like someone to discuss it.  :/

https://web2.0calc.com/questions/calculus-question_1

4) Inequality    Thanks Anonymous43338

https://web2.0calc.com/questions/mymaths

5) Function of a function.     Thanks Geno and Heureka

https://web2.0calc.com/questions/f-f-x-x-type-of-question#r2

6) Proof      Thanks Heureka

https://web2.0calc.com/questions/if-n-is-a-positive-integer-such-that-2n-1-is-a-perfect-square-show-that-n-1-is-the-sum-of-two-successive-perfect-squares#r1

7) Proof    Thanks guys but I would really like another mathematician to look at this one. :/

https://web2.0calc.com/questions/help_67408

@@ What is Happening?  [Wrap4]   Thurs 26/11/15   Sydney, Australia Time 9:38 pm   ♪ ♫

Hi all,

Our great answers have been supplied by CPhill, Heureka, Alan, Riddler, gibsonj338, Olivia3.141592654, Buubleman, Bertie, Anonymous4338, Steven434, Hayley, jc and guests.    Thank you.   

Interest Posts:

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.  

1)  Fun facts about 666     Thanks CPhill

     One has to wonder about Chris's motives LOL

Thank you Kitty <3         

What is the answer to square root 10 over 2 doubled?

\(2\times \frac{\sqrt{10}}{2}=\sqrt{10}\)