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Coin Flip

How Psychic Are You?

COINFLIP APPLET INSTRUCTIONS
Concentrate on the coin landing HEADS up.
Click the [Flip Coin] button to flip your coin.
Your coin flip is recorded in the lower right.
Repeat steps A to C ten times.
When finished, you will have a record of each flip, and
The total number of HEADS you successfully willed to occur.
This is the answer to question 9.
Do not cheat by continuing to play till you accumulate a humongous horde of heads.
There is a maximum number possible.
And there is no point in it.

ASSIGNMENT

EXPECTED PROBABILITY:
(assumes random guessing with no telekinetic ability)

In 10 flips, how many ways can you flip
- 0 heads
- 1 head
- 2 heads
- 3 heads
- 4 heads
- 5 heads
- 6 heads
- 7 heads
- 8 heads
- 9 heads
- 10 heads
Fill out the table below with
The results from question 1 in column C.
The probability of x heads in column P(x).
The product of columns x and P(x) in column x*P(x).
Sum each column on the SUMs row.
- What should the sum of the C column be?
- What should the sum of the P(x) column be?
- What does the sum of the x*P(x) column mean?
In 10 flips, how many times would you expect a coin to land HEADS up?
How many ways can you flip a coin 10 times?
(How many 10 coin sequences are there?)

Copy the table below into a comment.
Fill in the expected probabilities for each possible 10 coin flip.

EXPECTED PROBABILITY

x 0 1 2 3 4 5 6 7 8 9 10 SUMS
C
P(x)
x·P(x)

8. Write a formula for the probability of flipping x heads out of n flips.
Generalize your table above into a formula that inputs the number of heads x, the number of flips n,
and outputs the probability P of any number of heads in any number of flips.

Formula:	P(for x number of heads in n flips)
P(x) =

Your formula may have P(x) only on the left side. Your formula requires x and n on the right side.
Observe what you have done for specific x's. Then generalize your formula for any x.
Copy your table into a comment and fill in with your formula.

The OBSERVED DATA table asks these questions:

OBSERVED DATA:

Fill in a row of the table below with the observed data for your 10 coin flips.

P(#HEADS=x) is the probability of flipping your number of heads in 10 flips.
P(#HEADS£x) is the probability of flipping less than or the same number of heads.
P(#HEADS³x) is the probability of flipping more than or the same number of heads.

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?

Average So Far =>

Copy this table to a comment and fill in appropriately.

As a class, you may earn a group score of 8 points.
As individuals, you may earn another 6 points.
That's a big sum of 14 points for this assignment!
No points till you earn the last 6 points!
May the Force be with you as always...

Comments:

From TBird - 12/13/06 12:27 PM

EXPECTED PROBABILITY

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.0009756	0.009756	0.04395	0.1172	0.2051	0.2461	0.2051	0.1172	0.04395	0.009765	0.0009765	1.00
x·P(x)	0	0.009765	0.0879	0.3516	0.8204	1.2305	1.2306	0.8204	0.3516	0.087885	0.009765	5

In 10 flips, how many times would you expect a coin to land HEADS up?

In ten flips you would expect a coin to land HEADS up 5 times. 10*.5 = 5

How many ways can you flip a coin 10 times?
(How many 10 coin sequences are there?)

You can flip a coin ten times and get a total of 1024 sequences.
2*2*2*2*2*2*2*2*2*2 = 2^10 = 1024 sequences

Formula:	P(for x number of heads in n flips)
P(x) =	n(n-1)(n-2)…/((x(x-1)(x-2)…)((n-x)(n-x-1)(n-x-2)…)

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTHTHTTTH	5	0.2461	0.62308	0.62308	no
Average So Far =>	4	0.1172	0.37698	0.62308	no

From Bubba - 12/12/06 6:06 PM

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	.000977	.00977	.0439	.1172	.2051	.2461	.2051	.1172	.0439	.00977	.000977	1
x·P(x)	0	.00977	.0878	.3516	.8204	1.2305	1.2306	.8204	.3512	.08793	.00977	5

5. 1024

6. 50/50

7. 1024

Formula:	P(for x number of heads in n flips)
P(x) =	(nn-1n-2....)/(xx-1x-2...)(n-xn-x-1n-x-2...)/2^n

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
	5	.2461	.62308	.62308	noooooo
Average So Far =>	5	.2461	.62308	.62308	noooooo

I think we all need to stay away from Vegas, unless we want to go for the free drinks!

From Fro - 12/8/06 9:22 PM

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.000977	0.009766	0.043945	0.117188	0.205078	0.246094	0.205078	0.117188	0.043945	0.009766	0.000977	1
x·P(x)	0	0.009766	0.087891	0.351563	0.820313	1.230469	1.230469	0.820313	0.351563	0.087891	0.009766	5

5: 1,024 (2 coins, 10 flips, 2*2*2*2*2*2*2*2*2*2=1024.

6: 24.6%

7: 1024

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HTTHHTHHTT	5	.246094	.62308	.62308	NO
Average So Far =>	5	.246094	.62308	.62308	NO

From Kathi - 12/8/06 9:02 AM

I confused myself again and over thought it.

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	.000977	0.009766	0.043945	0.117188	0.205078	0.246094	0.205078	0.117188	0.043945	0.009766	.000977	1
x·P(x)	0	0.009766	0.087891	0.351563	0.820313	1.230469	1.230469	0.820313	0.351563	0.087891	0.009766	5

#5: The sum of row C is 1,024

Flipping 2 coins, 10 times,2*2*2*2*2*2*2*2*2*2=1024.

The sum of row P(x) is 1

The sum of x*P(x) is 5

#6: The highest probability is 5 times

#7: There are 1024 different combinations from 10 flips.

#8: P(x) = (n*n-1*n-2*n-...) / (x*x-1*x-2*x-...)*(n-x*n-x-1*n-x-2*n-x-...) / 2^n

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTTHHTTHT	5	0.246	0.623	0.623	NO
Average So Far =>	5

From Pringle - 12/7/06 1:56 PM

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.0009765	0.0097652	0.0439453	0.1171875	0.2050781	0.2460937	0.2050781	0.1171875	0.0439453	0.0097652	0.0009765	0.9999999=1
x·P(x)	0	0.0097656	0.0878906	0.3515625	0.8203125	1.2304687	1.2304687	0.8203125	0.3515625	0.0878906	0.0097656	4.999999=5

5. What should the sum of the C column be? 1024 (2*2*2*2*2*2*2*2*2*2=1024)

What should the sum of the P(x) column be? 1(total probability should be equal to 1)

What does the sum of the x*P(x) column mean? sum of the x*P(x)= 5 (when you flip the coin 10 times, 5 times will be heads up.)

6. In 10 flips, how many times would you expect a coin to land HEADS up? 5

7. How many ways can you flip a coin 10 times? 1024 (2^n=2^10=1024)
(How many 10 coin sequences are there?) flip a coin once= 2 ways (h) (t)

flip a coin twice= 4 ways (h,h)(h,t)(t,h)(t,t)

flip a coin 3 times= 8ways( h,h,h)(h,h,t)(h,t,h)(t,h,h)(h,t,t)(t,h,t)(t,t,h)(t,t,t)

Formula:	P(for x number of heads in n flips)
P(x) =	nn-1..* n-(n-1)/n-x* n-x-1..n-x-(x-1)/xx-1..x-(x-1)/2^n

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
hhthhhhhth	8	0.04394531	0.9892583	0.054687	no
Average So Far =>	5

From wHolt - 12/7/06 11:19 AM

Tiger - no points till you flip the coin and make a table of your observations.

From Tiger - 12/6/06 6:45 PM

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	.000977	.00977	.0439	.1172	.2051	.2461	.2051	.1172	.0439	.00977	.000977	1
x·P(x)	0	.00977	.0878	.3516	.8204	1.2305	1.2306	.8204	.3512	.08793	.00977	5

In 10 flips, how many times would you expect a coin to land heads up?

10*.5=5 or half the time

How many ways can you flip a coin 10 times? 1024

2 sides on a coin 10 times 2*2*2*2*2*2*2*2*2*2=1024

From wHolt - 12/6/06 12:27 PM

Cheana - you ignored the = equal part of questions 12 and 13

Kathi - why did you do this exercise twice?
is .5 the correct answer for #12 and #13?

it would be if the question asked for x<5 and x>5, but they dont.

From Kathi - 12/6/06 6:48 AM

EXPECTED PROBABILITY

x 0 1 2 3 4 5 6 7 8 9 10 SUMS
C 1 10 45 120 210 252 210 120 45 10 1 1024
P(x) .000977 0.009766 0.043945 0.117188 0.205078 0.246094 0.205078 0.117188 0.043945 0.009766 .000977 1
x*P(x) 0 0.009766 0.087891 0.351563 0.820313 1.230469 1.230469 0.820313 0.351563 0.087891 0.009766
5

#5: The sum of row C is 1,024

Flipping 2 coins, 10 times,2*2*2*2*2*2*2*2*2*2=1024.

The sum of row P(x) is 1

The sum of x*P(x) is 5

#6: The highest probability is 5 times (50% of the time it would land heads up).

#7: There are 1024 different combinations from 10 flips.

#8: P(x) = (n*n-1*n-2*n-...) / (x*x-1*x-2*x-...)*(n-x*n-x-1*n-x-2*n-x-...) / 2^n

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTTHHTTHT	5	50% or .5	0.5	0.5	NO
Average So Far =>	5	50% or .5	0.5	0.5	no

From wHolt - 12/6/06 12:22 AM

Cheana - you ignored the = equal part of questions 12 and 13

Kathi - why did you do this exercise twice?

From Houdini - 12/5/06 11:13 PM

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	.0009765625	.009765625	.0439453125	.1171875	.205078125	.24609375	.205078125	.1171875	.0439453125	.009765625	.0009765625	1
x·P(x)	0	.009765625	.087890625	.3515625	.8203125	1.23046875	1.23046875	.8203125	.3515625	.087890625	.009765625	5

5. The total number of possibilities of ten flips.

It should equal one.

It means that out of ten flips you should get five heads.

6. Five

7. 1024 ways.

Formula:	P(for x number of heads in n flips)
P(x) =	(n(n-1)(n-2)...)/((x(x-1)(x-2)...)((n-x)(n-x-1)(n-x-2)...))/2^n

Let's do 13 flips with 8 heads:

(13x12x11x10x9x8x7x6x5x4x3x2x1)/((8x7x6x5x4x3x2x1)(5x4x3x2x1))/2^13

6227020800/4838400/8192=0.1571044921875

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HTHTTTHTHH	5	.24609375	0.623046875	0.623046875	Nope
Average So Far =>	50%

I suck in Vegas, but hey they have free drinks!!!!

From Cheana - 12/5/06 10:57 PM

EXPECTED PROBABILITY

x 0 1 2 3 4 5 6 7 8 9 10 SUMS
C 1 10 45 120 210 252 210 120 45 10 1 1024
P(x) 9.765e-4 .0097 .0439 .1171 .20507 .2460 .20507 .1171 .0439 .0097 9.765e-4 .999493125
x·P(x) 0 .0097 .0878 .3513 .82028 1.23 1.230 .8197 .3512 .0873 .009765 4.997045

Formula:	P(for x number of heads in n flips)
P(x) =	n!/(x!(n-x)!)/ 2^n

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTHTTTHHT	5	50%	50%	50%	NO
Average So Far =>

From Melewen - 12/5/06 6:19 PM

EXPECTED PROBABILITY
x 0 1 2 3 4 5 6 7 8 9 10 SUMS
C 1 10 45 120 210 252 210 120 45 10 1 1024
P(x) .00097656 .0097656 .04395 .1171875
.205078 .24609375 .205078
.1171875
.04395
.0097656
.00097656
1
x*P(x)
0 .009656 .087890625 .3515625 .8203125 1.23046875 1.230468
.8203125
.3516
.0878904
.0097656
5

You can find your C column numbers by using a combination equation (10x9x8x7/4x3x2x1, etc).
For the probability of 2 you would do 10x9/2x1.

The sum of column C should be 1024, which is the total number of ways to flip the coin ten times.
The sum of P(x) should be 1, because all probabilities should add up to 1.
The sum of x*P(x) should be 5, because there's a 50/50 chance of getting a heads or tails on any given flip. And half of 10 is 5.

6. You would expect it to land heads up five times out ot ten, since the probability of getting a heads on any given flip is .5. .5*10 is 5.
7. 1024, the sum of all the individual possibilities to flip the coin, accounts for all the possible combinations.

Formula:	P(for x number of heads in n flips)
P(x) =

From wHolt - 12/4/06 9:52 AM

Taurus - dont pack up and move to Vegas yet.
You forgot to include the equals in the more than or equal to 7.
You figured only more than 7.
Notice what happens if you add the probability for exactly 7 to your calculation.

This goes for the rest of you also.
Dont forget to include the = in more than or less than your observed number of heads
for #12 and #13;
Or you may get a swollen head thinking you are psychic.

Kathi -
It is not the sum of x-P(x) , but x*P(x) that you are copying.
Also 5 out of 10 is not 25%.

From Kathi - 12/4/06 6:59 AM

EXPECTED PROBABILITY

x 0 1 2 3 4 5 6 7 8 9 10 SUMS
C 1 10 45 120 210 252 210 120 45 10 1 1024
P(x) .000977 0.009766 0.043945 0.117188 0.205078 0.246094 0.205078 0.117188 0.043945 0.009766 .000977 1
x·P(x) 0 0.009766 0.087891 0.351563 0.820313 1.230469 1.230469 0.820313 0.351563 0.087891 0.009766
5

#5: The sum of row C is 1,024

Flipping 2 coins, 10 times,2*2*2*2*2*2*2*2*2*2=1024.

The sum of row P(x) is 1

The sum of x-P(x) is 5

#6: The highest probability is 5 times (25% of the time it would land heads up).

#7: There are 1024 different combinations from 10 flips.

#8: P(x) = (n*n-1*n-2*n-...) / (x*x-1*x-2*x-...)*(n-x*n-x-1*n-x-2*n-x-...) / 2^n

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTTHHTTHT	5	0.246	0.623	0.623	NO
Average So Far =>	5	0.246	0.623	0.623	no

From Zonino - 12/3/06 11:16 PM

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.000977	0.00977	0.0439	0.117	0.205	0.246	0.205	0.117	0.0439	0.00977	0.000977	1
x·P(x)	0	0.00977	0.0878	0.351	0.82	1.23	1.23	0.819	0.3512	0.08793	0.00977	5

The sum of row C is the total number of possible flips.
The sum of P(x) should always be 1.
The row x*P(x) is the average that you should expect to get heads. Basically you should expect to get the coin to land on heads 5 out of 10 times or 50%.

6) You would expect the coin to land heads 50% of the time per the above, so out of ten flips, you should expect 5 heads.

7) You can flip each coin one of two ways, either a heads or a tails. So we would take the number of ways you can flip each coin and multiply them together. This is 2*2*2*2*2*2*2*2*2*2 or 1024 ways.

Formula:	P(for x number of heads in n flips)
P(x) =

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTTHHHTTT	5	0.246
Average So Far =>

From Taurus - 12/3/06 10:55 PM

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
hhhthhthht	7 (im lucky)	.1171	.82676	.05366	Yes! Vegas!
Average So Far =>	5.111

From Capricorn - 12/3/06 10:49 PM

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
TTHHTTHTHT	4	0.205078125	.17076	.622736	NO
Average So Far =>	4

From Capricorn - 12/3/06 10:26 PM

EXPECTED PROBABILITY

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.0009765625	0.009765625	0.0439453125	0.1171875	0.205078125	0.24609375	0.205078125	0.1171875	0.0439453125	0.009765625	0.0009765625	1
x·P(x)	0	0.009766	0.087891	0.351563	0.820313	1.230469	1.230469	0.820313	0.351563	0.087891	0.009766	5

5: 1,024 (2 coins, 10 flips, 2*2*2*2*2*2*2*2*2*2=1024.

6: 5 heads. 10 flips times the probability of getting a head (or tails) which is 0.5 =5 heads or 5 tails

7: 1024 as well. You can flip the coin and get many different combinations and ways it can turn out.

8: I also agree with Pac

Formula:	P(for x number of heads in n flips)
P(x) =	(nn-1n-2....)/(xx-1x-2...)(n-xn-x-1n-x-2...)/2^n

From Taurus - 12/3/06 10:22 PM

EXPECTED PROBABILITY

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.0009765625	0.009765625	0.0439453125	0.1171875	0.205078125	0.24609375	0.205078125	0.1171875	0.0439453125	0.009765625	0.0009765625	1
x·P(x)	0	0.009766	0.087891	0.351563	0.820313	1.230469	1.230469	0.820313	0.351563	0.087891	0.009766	5

5: The sum of row C should be 1,024 (2 coins, 10 flips, 2*2*2*2*2*2*2*2*2*2=1024.

6: 10 x 0.50 = 5 Heads There is .5 probability that will a Head or Tail will land. So 10 flips times the probability of .5 is 5 heads.

7: 1024, 2^10. 2 ways every flip.

8: I'm going to have to agree with Pac on this one

Formula:	P(for x number of heads in n flips)
P(x) =	(nn-1n-2....)/(xx-1x-2...)(n-xn-x-1n-x-2...)/2^n

From 7Iron - 12/3/06 7:55 PM

EXPECTED PROBABILITY

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.0009756	0.009756	0.04395	0.1172	0.2051	0.2461	0.2051	0.1172	0.04395	0.009765	0.0009765	1.00
x·P(x)	0	0.009765	0.0879	0.3516	0.8204	1.2305	1.2306	0.8204	0.3516	0.087885	0.009765	5

In 10 flips, how many times would you expect a coin to land HEADS up?

In ten flips you would expect a coin to land HEADS up 5 times. 10*.5 = 5

How many ways can you flip a coin 10 times?
(How many 10 coin sequences are there?)

You can flip a coin ten times and get a total of 1024 sequences.
2*2*2*2*2*2*2*2*2*2 = 2^10 = 1024 sequences

Formula:	P(for x number of heads in n flips)
P(x) =	n(n-1)(n-2)…/((x(x-1)(x-2)…)((n-x)(n-x-1)(n-x-2)…)

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTHTHTTTH	5	0.2461	0.62308	0.62308	no
Average So Far =>	4	0.1172	0.37698	0.62308	no

*I am only using one entry/classmate for my average 4+4+5 = 13
13/3 = 4 1/3 = 4

From Bubba - 12/3/06 7:28 PM

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	.000977	.00977	.0439	.1172	.2051	.2461	.2051	.1172	.0439	.00977	.000977	1
x·P(x)	0	.00977	.0878	.3516	.8204	1.2305	1.2306	.8204	.3512	.08793	.00977	5

In 10 flips, how many times would you expect a coin to land HEADS up? half

How many ways can you flip a coin 10 times?
(How many 10 coin sequences are there?)

From GolfGirl - 12/3/06 3:02 PM

EXPECTED PROBABILITY

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.000977	0.009766	0.043945	0.117188	0.205078	0.246094	0.205078	0.117188	0.043945	0.009766	0.000977	1
x·P(x)	0	0.009766	0.087891	0.351563	0.820313	1.230469	1.230469	0.820313	0.351563	0.087891	0.009766	5

5: The sum of row C should be 1,024 (2 coins, 10 flips, 2*2*2*2*2*2*2*2*2*2=1024. The sum of x*P(x) is 5, which means that ....

6: The highest probability will be 5 times 24.6% of the time it would land heads up.

7: There are 1024 different combinations of heads or tails

Formula:	P(for x number of heads in n flips)
P(x) =	n*.5

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHHTHHTHTT	6	.2051	.8281	.3769	NO
Average So Far =>	4	.2050781	.7724607	.8281248	NO

From wHolt - 12/3/06 1:36 PM

Boki - when I add up your probabilities less than and including 3,
I get a number bigger than .10 .
[The chance of P(x=3) = .11 > .10]
Rethink what less than or equal mean,
and also what more than or equal mean.

From Harkar - 12/2/06 5:42 PM

EXPECTED PROBABILITY

x 0 1 2 3 4 5 6 7 8 9 10 SUMS
C 1 10 45 120 210 252 210 120 45 10 1 1024
P(x) .0009765 .0097656 .0439453 .1171875 .2050781 .2460937 .2050781 .1171875 .0439453 .0097656 .0009765 1
x·P(x) 0 .0097656 .0878906 .3515625 .8203124 1.2304685 1.2304686 .8203125 .3515624 .0878904 .009765 5

Row C Explained:
0 head   - 10*9*8*7*6*5*4*3*2/ 10*9*8*7*6*5*4*3*2*1=1
1 head   - 10/1=10
2 heads  - 10*9/ 2*1=45
3 heads  - 10*9*8/ 3*2*1=120
4 heads  - 10*9*8*7/ 4*3*2*1=210
5 heads  - 10*9*8*7*6/ 5*4*3*2*1=252
6 heads  - 10*9*8*7*6*5/ 6*5*4*3*2*1=210
7 heads  - 10*9*8*7*6*5*4/ 7*6*5*4*3*2*1 =120
8 heads  - 10*9*8*7*6*5*4*3/ 8*7*6*5*4*3*2*1 =45
9 heads  - 10*9*8*7*6*5*4*3*2/ 9*8*7*6*5*4*3*2*1 =10
10 heads - 10*9*8*7*6*5*4*3*2/ 10*9*8*7*6*5*4*3*2*1 =1

5.
a. The sum of row C is 1,024 (2 coins and 10 flips= 2*2*2*2*2*2*2*2*2*2).
b. The sum of row P(x) should be 1 because the sum of probabilities is always 1.
c. The sum of x*P(x) is 5 which is the number of heads in 10 flips times the probability of getting that number of heads – which means that in 10 flips you would expect to have 5 heads. 6. 5 times

7.1024

Formula:	P(for x number of heads in n flips)
P(x) =

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
TTTTTHTHHH	4	.2050781 = 20.51%	.7724607 = 77.25%	.8281248 = 82.81%	NO
Average So Far =>	4.3 = 4	.2050781 = 20.51%	.7724607 = 77.25%	.8281248 = 82.81%	NO

From Boki - 12/2/06 5:19 PM

less than 10% sholud be "less then 0.1"

line 13 is:

0.00259 <.1

.00156<.1

and so on.. and answer should be YES as I did it before.

From Pac - 12/2/06 2:00 PM

EXPECTED PROBABILITY

#1-4:

x	0	1	2	3	4	5	6	7	8	9	10	SUMS
C	1	10	45	120	210	252	210	120	45	10	1	1024
P(x)	0.000977	0.009766	0.043945	0.117188	0.205078	0.246094	0.205078	0.117188	0.043945	0.009766	0.000977	1
x·P(x)	0	0.009766	0.087891	0.351563	0.820313	1.230469	1.230469	0.820313	0.351563	0.087891	0.009766	5

#5: The sum of row C should be 1,024 (2 coins, 10 flips, 2*2*2*2*2*2*2*2*2*2=1024. The sum of row P(x) should be 1 because it is the sum of probabilities and there really should be 100% of probabilities accounted for. The sum of x*P(x) is 5, which means that

#6: The highest probability would be 5 times (24.6% of the time it would land heads up). But, if you widened the options and said that between 4-6 times it will land heads up, the probability increases to 65.6%.

#7: There are 1024 different combinations of heads/tails that could result from 10 flips of a coin.

#8:

Formula:	P(for x number of heads in n flips)
P(x) =	(nn-1n-2n-...) / (xx-1x-2x-...)(n-xn-x-1n-x-2n-x-...) / 2^n

Example: if n=10 and x=5, we already know that P(x) is 0.246094 from the table in the first part of the assignment. In the formula: (10*9*8*7*6*5*4*3*2*1) / (5*4*3*2*1)*(5*4*3*2*1) / 2^10 = (3628800) / (120)*(120) / 1024 = 3628800/14400/1024=0.24609375

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HTTHHTTTTT	3	0.117188	0.171875	0.945313	Nope
Average So Far =>	4.3333

My average of 4.3333 includes each of Boki's sequences separately for a total of 9 trials including Draco's and my own.

From wHolt - 12/2/06 12:42 AM

Boki - If line 13 is less than 10%, enter YES, otherwise NO.
What does less than 10% mean?

From Boki - 12/1/06 3:25 PM

NO, because each number in a row #13 is greater then 10%.

From wHolt - 12/1/06 1:56 PM

Boki - Are you psychic if your highest number of heads is 5?

Draco -
Since n=10, your formula says P(x) = 10*.5 = 5.
But P(x) is not the same for x=3 or x=6, etc.
P(x) is not always 10*.5=5 .
What does your formula predict?

Also you will need to do some adding or subtracting for answers #12 and #13

From Draco - 12/1/06 12:46 AM

EXPECTED PROBABILITY

x 0 1 2 3 4 5 6 7 8 9 10 SUMS
C 1 10 45 120 210 252 210 120 45 10 1 1024
P(x) .000977 or .098% .00977 or .98% .0439 or 4.39% .1172 or 11.72% .2051 or 20.51% .2461 or 24.61% .2051 or 20.51% .1172 or 11.72% .0439 or 4.39% .00977 or .98% .000977 or .098% 1 or 100%
x·P(x) 0 .00977 .0878 .3516 .8204 1.2305 1.2306 .8204 .3512 .08793 .00977 5

Formula:	P(for x number of heads in n flips)
P(x) =	n*.5

Basically there is a 50% chance for the coin to land on either heads or tails.

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
HHTHHHTTTH	6	.2051 or 20.51%	.8281 or 82.81%	.3769 or 37.69%	NO
Average So Far =>

From wHolt - 11/29/06 1:33 PM

Boki - nice table, but we cannot see how you got your numbers.
Don't use any notation we cannot understand. C(10,10)?
Relate all to previous assignments.

From Boki - 11/29/06 12:38 AM

In 10 flips, how many ways can you flip

With ten flips there are 2^10=2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 1024 possibilities.

In order to find how many ways can you flip 0, 1, 2,.... 10 heads in 10 flips, we can use this formula for combinations:

n * (n-1) * (n-2) * ... * (n - k + 1)

C(n,k) = -------------------------------------------- = n/k*(n-1)/(k-1)*(n-2)/ (k-2)*………*(n-k+1)/1

k*(k-1)*(k-2)*………* 1

0 heads       C(10,0) =10*9*8*7*6*5*4*3*2/1*10*9*8*7*6*5*4*3*2=1

1 head          10/1=10
2 heads         10*9/2=45
3 heads         10*9*8/6=120
4 heads         10*9*8*7/24=210
5 heads         10*9*8*7*6/120=252
6 heads         10*9*8*7*6*5/720=210
7 heads         10*9*8*7*6*5*4/5040 =120
8 heads         10*9*8*7*6*5*4*3/40320 =45
9 heads         10*9*8*7*6*5*4*3*2/362880 =10

10 heads 10*9*8*7*6*5*4*3*2/3628800 =1
EXPECTED PROBABILITY
x
0
1
2
3
4
5
6
7
8
9
10
SUMS
C
1
10
45
120
210
252
210
120
45
10
1
1024
P(x)
0.0009765625
0.009765625
0.0439453125
0.1171875
0.205078125
0.24609375
0.205078125
0.1171875
0.0439453125
0.009765625
0.0009765625
1
x·P(x)
0
0.009765625
0.087890625
0.3515625
0.8203125
1.23046875
1.23046875
0.8203125
0.3515625
0.087890625
0.009765625
5
6.In 10 flips, how many times would you expect a coin to land HEADS up?
10 x 0.50 = 5 Heads (there is .5 probability that will either Head or Tail land, so 5 times the coin will land heads up or tails up)
7.How many ways can you flip a coin 10 times?
in ten flips there is only 1 way to flip a coin 10 times

(How many 10 coin sequences are there?)
there is only 1 ten-coin sequence
8. Write a formula for the probability of flipping x heads out of n flips.
Generalize your table above into a formula that inputs the number of heads x, the number of flips n,
and outputs the probability P of any number of heads in any number of flips.
Formula:
P(for x number of heads in n flips)
P(x) =
n*.5

OBSERVED DATA

#9	#10	#11	#12	#13	#14
SEQUENCE	#HEADS	P(#HEADS=x)	P(#HEADS£x)	P(#HEADS³x)	PSYCHIC?
TTHHTHTHHT	5	0.24609375	0.0015673981191222570532915360501567 0.15>.10	0.0025906735751295336787564766839378 0.25>.10	NO
TTHHTTTTHH	4	0.205078125	0.0025906735751295336787564766839378 0.25>.10	0.0015673981191222570532915360501567 0.15>.10	NO
HTHHTTHHTT	5	0.24609375	0.0015673981191222570532915360501567 0.15>.10	0.0025906735751295336787564766839378 0.25>.10	NO
HTTTHHTHHT	5	0.24609375	0.0015673981191222570532915360501567 0.15>.10	0.0025906735751295336787564766839378 0.25>.10	NO
THTTHHTHTT	4	0.205078125	0.0025906735751295336787564766839378 0.25>.10	0.0015673981191222570532915360501567 0.15>.10	NO
TTTTHTHTHT	3	0.1171875	0.0056818181818181818181818181818182 0.56>.10	0.0011792452830188679245283018867925 0.11>.10	NO
HTTHTTTTHH	4	0.205078125	0.0025906735751295336787564766839378 0.25>.10	0.0015673981191222570532915360501567 0.15>.10	NO
Average So Far =>	4.3	0.2101005	0.0025937190377962220020465509114286 0.25>.10 even everyone is greater then 10%, stay away from Vegas ANYWAY!	0.001950494337967748588667477155 0.19>.10	NO

#12: P(#HEADS£x)

x=5
C(10,0)+C(10,1)+C(10,2)+C(10,3) +C(10,4)+C(10,5)= 638

P=1/ 638=0.0015673981191222570532915360501567

x=4

C(10,0)+C(10,1)+C(10,2)+C(10,3) +C(10,4)=1/386

P=1/386=0.0025906735751295336787564766839378

x=3

C(10,0)+C(10,1)+C(10,2)+C(10,3)=176
P=1/176= 0.0056818181818181818181818181818182

#13: P(#HEADS³x)

x=5
C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)= 210+120 +45+10+1=386

P=1/386=0.0025906735751295336787564766839378

x=4
C(10,5)+ C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)= 638

P=1/ 638=0.0015673981191222570532915360501567

x=3

C(10,4)+C(10,5)+ C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)= 848

1/848= 0.0011792452830188679245283018867925

From wHolt - 11/28/06 10:37 AM

Dont miss out on these last 4 problems. They alone will raise your grade by one level.

Last Modified 12/7/06 11:21 AM

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