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Is this really your calculation? It looks like you copied it from somewhere.

If it is yours, then how did you formulate your calculation?

It is unlikely what you presented is the correct probablity.

Re-Present your calculation as an equation with the work product: delineate your calculations showing how the Inclusion-Exclusion principle applies to your calculation. Specifically, define the included set, the excluded set, and the union of these two sets. 

If you do this, then it should be easy for someone to verify your solution, without deriving and solving this. (As an analogy, it’s easy to verify the primes of large numbers; it’s not so easy to find the primes in the first place.) 

If you do decide to do this, use LaTex. Reading math in ASCII slop is often irritating.

GA

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Also, I posted this last night but didn't got any responses so I hope someone can help me cuz I need the help asap(It's due today)

You’re an idiot waiting until the last day to ask for help for a question this complex and laborious.

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Top   8 x 2 rectangle =   16 cm^2

Middle       5 x ( 9 + 8) rectangle   =  5  x 17 =   85 cm^2

Bottom   right triangle with a height of 5  and a  base of ( 17 - 6 - 3)  =  8

Area =  (1/) 8 * 5 =  20 cm^2

Total area = 

16 + 85 + 20  = 

121 cm^2

Top - half circle with a radius of  (6 -2 -1) / 2 =  3/2  =1.5

Area =  3.14 (1.5)^2

Middle  - a rectangle  with dimensions of  (6 + 1) x  4 =  7 x 4 =  28

Bottom - a rectangle with dimensions  of   6 x 3  = 18

Total area  =

 [ 3.14 * 1.5^2   + 28 + 18 ]  km^2  =

53.065  km^2

Top left - We have a half circle with a radius of ( 8 - 4 + 2)  / 2 = 6/2 = 3     area =     3.14 * 2^2 =  12.56   km^2

Center - We have  an 8  x 2 rectangle  =  16   km^2

And, at the bottom  we have a   5 x 8  rectangle = 40  km^2

Total area = 

12.56 + 16 + 40 = 

68 .56    km^2

I believe there is a formula for this that  is similar to Heron's 

s =  (  15 + 36 + 39) /  2  =    90/ 2  = 45

Area =  (4/3)  sqrt   [ s ( s - m1) (s - m2) (s - m3) ] = 

(4/3)  sqrt [ 45 * 30 * 9 * 6 ]   =  

360 units^2

70 + 3x  =  80 + 2x

3x - 2x =  80  - 70

x  = 10    hours

Here's the graph :  https://www.desmos.com/calculator/lxml7xtmnl

Cost of hot dog meal  =  x

Cost of hamburger meal  = y

2x + 1y =  18     →    y  =   18 - 2x      (1)

3x  + 2y =  31     (2)

Sub (1) into (2)

3x + 2 ( 18 - 2x) =  31

3x + 36  - 4x  =  31

3x - 4x  =  31  - 36

-x =  -5

x = 5  = cost (in dollars)  of hot dog meal

Hamburger meal  costs   18 - 2(5) =   18 - 10  =  $8

See my answer under your original post

Call the nymber of friends he has, x

5x + 25  =   the total number of cupcakes  he had before baking the additional ones

Giving his friends 9 cupcakes apiece  would result in this equation

5x +  25 + 35   =  9x

5x + 60 =    9x

60   = 9x - 5x

60  = 4x

60 / 4  = x =    15  friends

The number he has baked =  5 *x + 25 =   5(15) + 25  =  100

Proof

If he bakes 35 more he has 135

And each friend would receive     135 / 15  =   9

To have two real roots, the discriminant must be >  0

So

m^2  -  4(2)(18)  >  0

m^2  >  144

This  will be true  when   m  =  ( -inf - 12)U ( 12, inf)

O + J  =  66    

O -  J =   24         add these

2O  =     90

O  = 90 / 2   =  $45  = what Oliver had  at all times

J =  $21  = what Juan had at first   

Juan had  (1/3)O  =   (1/3) (45)   =  $ 15  after the store 

He spent  21 - 15  =       $ 6 at the store

It is a power of 4 so there are 4 solutions

this is one way to tackle it
\(x^4=-64\\ x^2=8i\qquad or \qquad x^2=-8i\\ x=\pm\sqrt 8 [\sqrt i] \qquad or \qquad x \pm\sqrt 8\sqrt{-1} [\sqrt i] \\ x=\pm\sqrt 8 [\sqrt i] \qquad or \qquad x \pm i\sqrt 8 [\sqrt i] \\ x=\pm2\sqrt 2 [\sqrt i] \qquad or \qquad x \pm2 i\sqrt 2 [\sqrt i] \\ \)

Now you need to check what I have done and work out what the square root of  i  is.

there is lots of places on the internet help you to learn how to work this out.

Here is one:

Since BY = BC

Let  angles BCY and BYC = x = angle BCA 

Then angle YBC =  180  - 2x

And angle XBY =   150 - (180 - 2x) =  2x - 30 = Angle BXY

And since AX = XY

The angles  XAY and XYA are equal

And by the exterior angle theorem, angle BXY =  2* angle  XAY= angle BAC 

2 * angle BAC = angle BXY

2 * angle BAC =   2x - 30

angle BAC = (2x - 30) / 2 =  x - 15

So

angle BAC +  angle ABC  + angle  BCA = 180

(x - 15)   +  ( 150 )  +  (x)  =  180

2x +  135  = 180

2x =  180 - 135

2x  = 45

x = 22.5

Angle BAC =  22.5 - 15 =    7.5°

Ronnie had R stamps at first

Harry  had  6R

After  Harry gave  1/2 of his stamps to Ronnie

(1/2) * 6R  + 11  =  R + (1/2) (6R)

3R + 11  =  R + 3R

3R + 11 =  4R

11 =  4R - 3R

11 = R

Harry had 6*11 =    66 stamps at first

Note that a, b   are both positive

Using some log properties

log ( a - 2b)^2  =  log (a * b)

This means that

(a - 2b)^2  =  ab

a^2 - 5ab + 4b^2  = 0

(a - 4b) ( a - b)  = 0

Either                                          

a - b=   0        

a =  b

a / b  =  1      { this may produce a log of a  negative  - not  allowed }

Or

a - 4b  = 0

a = 4b

a / b =  4

The second is  correct

11 , 22 , 33 , 44 , 55 , 66 , 77 , 88 , 99 , 110 , 121 , 132 , 143 , 154 , 165 , 176 , 187 , 198 , 209 , 220 , 231 , 242 , 253 , 264 , 275 , 286 , 297 , 308 , 319 , 330 , 341 , 352 , 363 , 374 , 385 , 396 , 407 , 418 , 429 , 440 , 451 , 462 , 473 , 484 , 495 , 506 , 517 , 528 , 539 , 550 , 561 , 572 , 583 , 594 , 605 , 616 , 627 , 638 , 649 , 660 , 671 , 682 , 693 , 704 , 715 , 726 , 737 , 748 , 759 , 770 , 781 , 792 , 803 , 814 , 825 , 836 , 847 , 858 , 869 , 880 , 891 , 902 , 913 , 924 , 935 , 946 , 957 , 968 , 979 , 990 , Total =  90 such numbers.

The solutions are 1 + sqrt(3)*i, 1 - sqrt(3)*i, -1 + sqrt(3)*i, -1 - sqrt(3)*i

Answer by computer program

for (n = 1..1375)

  if (terminate(n/1375) == 1) {count = count + 1}

output(count)

output = 147

2x^2 = - 2y^2 + 12x - 4y + 20 + y^2 + x^2

Rearrange as

x^2 - 12x + y^2 - 4y   =  20     complete the square on x,y

x^2 -12x + 36  + y^2 - 4y + 4 =  20 + 36 + 4

(x - 6)^2  + (y - 2)^2  =   60

The radius of the  circle =  sqrt (60)

So  the side of the  square is  twice this =  2sqrt (60)

Area of the square =   [ 2sqrt (60) ] ^2  =  4 * 60  =   240

g(-1) =  -1 + 8 =  7

f (g(-1))  = f(7)  =  5(7) -  10  =   25

PQ = 12 - 3 =  9

That's incorrect

                                        B

                                                D                   

                             2

                 A                                           C

Note that angles BAD and DAC  = 15°

By the Law of Sines

sin BDA / 2  =  sin BAD / BD     

sin BDA / 2  = sin 15°/ BD

sin BDA =    2sin 15° / BD        (1)

Since AD bisects BAC, then 

AC / DC =  AB/ BD →  AC / 3BD = 2 /BD   →   AC =  3BD * 2 / BD =  6

And 

sin CDA = sin  BDA    so

sin CDA / 6 =  sin DAC / 3BD

sin BDA / 6 =  sin 15°  / 3BD   →  sin BDA = 6sin 15° / 3BD     (2)

Equating (1) and (2)

2 sin 15° / BD  =   6sin 15° / 3BD

3BD / BD =  6sin 15° / 2sin  15°

So

BC =   3BD + BD

BC^2  = (3B  + BD)^2

BC^2 =   ( 6sin 15° + 2sin 15°)^2   =   

 (sin 15° ( 6 + 2) )^2  =

sin^2 (15°) * 8^2  =

(sqrt 3 - 1)^2 / 8 * 64  =

(3 - 2sqrt 3 + 1) * 8  =

(4 - 2sqrt (3) ) * 8  = 

32 - 16sqrt (3)

p = 32   q = -16     r =   3

p + q + r  =   19