Answer to basketball simulation

If I set up my simulation like this:

01-72 = makes shot
73-00 = misses

here's some digits:

5730 3485 3246 75 4563
2pts 1pt 2pts 0pts 2pts.

For the 5 runs above my average would be: (2 + 1 + 2 + 0 + 2)/5 = 1.4

That's the simulation.

Now for the expected value

P(0) = 0.28 (misses first shot and is done)
P(1) = (.72)*(.28) (makes the first, misses the second)
P(2) = (.72)^2 (makes both)

Now just run the expected value formula with 0, 1, 2 and the 3 probabilities.

Answers, comments and hints to the review

3a) Note that the problem says that the tickets are bought at the same time for the same price. This makes the formula 3C + 5F.

If the each ticket was bought separately, that would make the formula C + C + C + F + F + ... The mean is the same for either formula. But if the second formula was used, you'd have to do the Pythagorean formula thing.

3b) First do 3C = 450 and 5F = 500; then do sqrt(450^2 + 500^2)

3c) This is only about ONE of each tix: C - F. You can take it from there...

6) Make a 2-way table: 51% in the upper left corner. The other two numbers go on the outside.

a) only 3% is left in the lower right corner
b) P(left|right) = 51/82 = 62.2%. Since P(left) = 66%, this is fairly different, so they are not independent.

7a) 1 - (89/90)^10
7b) 1 - (9/10)^10
7c) 1 - (89/90)^5*(9/10)^5

25d) (0.93)^4*(0.07)

28a) mean = 4

28b) sd = 3.2

28c) Think and read carefully! If the first is bigger than the second then:

(first - second) > zero

So we want to use the mean and sd from above to compare to zero:
z = (0 - 4)/3.2 = -1.249
P(z>-1.249) = (using normalcdf) = 89.4%

42a) 1/100 = 0.01

42b) (.99)(.99)(.01) = 0.009801

42c) (.99)^100 = 0.366

42d) You want to be first!


42e) It doesn't matter! Everyone has a 1% chance!

Now if you're thinking carefully about (e), you might be thinking: "Hey, don't the probabilities change?" Watch this!

Prob(3rd person wins) = (99/100)*(98/99)*(1/98) = 1% (notice how all the fractions reduce!)
Pretty cool, huh?

I will check e-mail at about 10-ish tonight. If you are feeling frustrated, drop a line to:

mrmathman @ gmail.com
I will reply tonight.

Good luck!

Review Answers

It is 8:20pm and I'm starting to put up some answers. Here they come...

THE LAST ASSIGNMENT OF 07!!!

Unit IV Review #3, 6, 7, 23, 25, 28, 42
Ch. 11 #15:  Run 10 times and find the expected value of this problem in theory.

07 is finishing...

Friday:  Ch. 8 #25, 35;  Ch. 10 #1
Monday:  Ch. 11 #2, 5, 19

Review

Wednesday:  we did a 2-page green worksheet that is an excellent probability review.
Thursday:  Unit I review; Page 106 #8, 15, 16, 30

Rules!

Ch. 16 #24, 28

Answers:
24a)  60, 12
b) 6, 1.5
c) 92, 12.37
d) 68, 12.37
e) X + X = 160, 16.97

28a) 10
b) 2.68
c) The seed packets need to be independent of each other.  If they all came from the same batch, they might not be independent.

The end of Unit 07

No homework tonight.

Review

Unit IV Review #1, 5a, 15ab, 16ac, 35-37

Even Answers:
16a) 38.9%
16c) 79.3%

36a) 59.29%
36b) 40.71%
36c) 5.29%
36d) 0.25%

Thursday

Article #7

More Expected Value

Ch. 16 #3, 5, 21
Scramble #2

Expected Value

Ch. 16 #16, 17

Quiz day

Scramble #1 for homework