CPhill

I'll take a stab at this one.....if the number of dice showing a two-digit number  = the number of dice showing a one-digit number, then  2 must show a two-digit number and 2  must show a one-digit number

There are 3 two-digit numbers and 9 one-digit numbers on each die

So....the probabiity is

C(4,2) (3/12)^2 (9/12)^2    =  

C(4,2) (1/4)^2 (3/4)^2  =

27/128 

l 6 - x  l  = 2

We actually have two equations here

6 - x  = 2               and              6 - x  =  -2

                           subtract 6 from both sides

-x  = - 4                                      -x   =  -8

                          multiply through by  -1

x  = 4                                           x  = 8

Thanks, hectictar.....!!!!

Yeah....GYangg...I just decided to expand it...but..we can see it this way.....

C(6,3)  (6x)^3 (1 / (3x))^3

Notice that the  x^3  terms will cancel   and we will be left with

C(6,3) * 6^3  / 3^3  =

20 * 216 / 27  

20 * 8  = 

160

Well...let's see!!!

(6x  +  1/ (3x) )^6   =

(6x)^6  +  6 * (6x)^5 (1/ (3x) )  +  15* (6x)^4 (1/(3x))^2  + 20* (6x)^3*(1 /(3x))^3

+ 15(6x)^2 * (1/(3x))^4 +   6* (6x)(1/(3x) )^5  + (1/ (3x))^6

The constant term is in red  

20 * 216x^3 / (27x^3)  =  20 *216 / 27   =  20 *24 / 3  =  20 * 8   =  160

13/ 32.5    =   2/5

I believe you are describing a "flush".....5 cards of the same suit

We  have  three types

Royal Flush  - 4

StraightFlush - 36

Flush  - 5108

Probability  =   [ 4 + 36 + 5108 ]  /  C (52,5)  ≈   0.00198  ≈  .198%

14....Mutually exclusive.....means that two things cannot happen simultaneously....so....you can't roll a 3 and a 4  at the same time...

12. 

P ( Working Radio  l Ready for Service )  = 

P ( Working Radio and Ready for Service)

__________________________________

           P(Ready for Service )

                                                                 =     .67

                                                                        ___

                                                                        .84

                                                                  =     67

                                                                        ___          

                                                                         84

                                                                  =   .7976  

                                                                    =  79.8 %  

13.   Here's my best attempt at this

We have  11  students who are honor students....so  40 - 11 =  29 aren't

We have 10 students who are athletes....but 3 of these are also  honor students

So....7  are athletes  but not honor students

So...the probability of  being an athlete given that the student is not an honor student  is

7/29   =  .241

10.  P (Brown or Dress)  =  P(Brown) + P(Dress)  - P (Brown and Dress) 

                                       =   (13/21)  + (10/21)  - ( 7/21)

                                      =    16/21 

11.  P ( More than 1 hour of TV  l 6th period )  =  13/23  = .565