CPhill

Picking 5 numbers from 41 can be done in C(41,5) = 749,398 ways

And choosing another number from 31 can be done in 31 ways.....so....

749,398 x 31  = 23,231,338  ways

So  1 / 23,231,338  = 0.0000000430453037  = 0.00000430453037% chance of winning on one ticket.....not too good, is it ???

Nice, MG1  !!!

 4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2

Rearanging both equations, we have

x = 8/(5- y)^2      and  x = 2/(3-y)^2

This implies that

 8/(5- y)^2  =  2/(3-y)^2      divide both sides by 2

4/(5 - y)^2  = 1/(3-y)^2       cross-multiply  and simplify

4(y^2 - 6y + 9) = y^2 - 10y + 25

4y^2 - 24y + 36  = y^2 - 10y + 25

3y^2 - 14y + 11  = 0    factor

(3y -11) (y - 1) = 0     set each factor to 0

y = 11/3    and y  = 1

And when y = 11/3, x =  9/2        

And when y = 1, x = 1/2

Here's the graph of both    x = 8/(5- y)^2      and  x = 2/(3-y)^2   and the points of intersection

f(x) = e^(2x) - e^(-x)

f ' (x) = 2e^(2x) + e^(-x)

I assume you're sayng that all the cars differ in at least one aspect......so...there are 24 "different" cars, thusly:

(H, W,2)          (V, W,2)            (F, W, 2)          (C, W,2) 

(H, W,4)          (V, W,4)            (F, W, 4)          (C, W,4)                   

(H, R, 2)          (V, R,2)             (F,  R,  2)         (C, R, 2)

(H, R, 4)          (V, R 4)             (F, R,  4)          (C, R, 4)  

Notice that there are 4 ways  to  pair the  Hondas with a "different" Volvo, "different" Ford  or "different" Cadillac.  So, 4 + 4 + 4 = 12

And there are 4 ways  to pair the Volvos wth a "different" Ford or "different' Cadillac. {We've already paired each with a "different" Honda.} So 4 + 4 = 8

And since we've already paired the Fords with a "different" Honda or Volvo, we have 4 ways that we can pair them with a "different" Cadillac.

So there are 12 + 8 + 4 = 24 "different" pairings

And the total number of possible pairs made by selecting any 2 cars from 24  = C(24,2) = 276

So 24 / 276 = 2/23 = about an 8.7% chance of pairing two "different" cars......

The angle of depression is given by...

tan^-1(10,000/ 50,000)   =  tan^-1(1 / 5) = about 11.31°

NO!!!    { All we need is one counter-example........ }

√7 + √7  ≠   2 (7)      prove this for yourself.....

Probability of a straight.....There are 10 ways to do this within the ranks. {The ace can be the high or low card.} And within each rank, we are choosing any 1 of 4 cards.  Note that we are including the straight flushes and royal flushes, here.

So.....there are C(10,1)*[C(4,1)]^5  = 10,240 possibilities

For the flush, we want to choose any 5 of the 13 ranks. And we have 4 ways to make each of these flushes.....all diamonds, all spades, all hearts or all clubs.  Here again, we are including the royal and straight flushes.

So......there are  C(13, 5) * C(4,1) = 5148 possibilities

Thus....the straight is almost twice as likely to occur......

Let A be the probability of drawing a white ball first, then a red one second, and then a white one last.

Let B be the probability of drawing a red ball first, then a white one second, and then a red one last.

So, P(A or B)  = 

[4/9 x 5/8 x 3/7 =  60/504 ] + [ 5/9 x 4/8 x 4/7 = 80/504 ]  =

[60 + 80 ] / 504  = 140/504 =  5/18 = about 27.8%