Question

In a game of dice, you roll two standard six-sided dice and examine their sum.
a) What is the probability of rolling a 7? 1/6
b) If you roll the dice 20 times, what is the probability of rolling a total of 3 sums of 7?
c) How many times would you expect to roll the dice before you rolled a sum of 12? 36

I already answered a and c, but I need help with b. If anyone could help, that'd be awesome. Thanks!

Answer 1

@beccaboo333

Answer 2

Yeah nope. Becca=no good math

Answer 3

Thanks anyway.

Answer 4

@Hero can you help?

Answer 5

@satellite73 can you help?

Answer 6

Can anyone help me? I have a lot to do and I really need to get through this assignment but I need help.

Answer 7

have a look at this chart

http://bestcase.files.wordpress.com/2011/01/dicediagram.jpg

it shows all 36 ways to roll 2 dice

Answer 8

the cells represent the sums of the two dice (from the row & and column headers)

Answer 9

oh wait, you already have the answer of 1/6, nvm

but you would could the number of 7s that show up (6 of them) and divide by 36 to get 6/36 = 1/6

So part a) is correct

Answer 10

Alright what about b?

Answer 11

to answer part b, we need to use a binomial distribution

Answer 12

in this case,

probability of success (ie rolling a 7) is p = 1/6
n = 20 (ie we have 20 trials or rolls)
k = 3 (we want exactly 3 rolls of 7)

and we plug all that into

(n C k)*(p)^(k)*(1-p)^(n-k)

Answer 13

the notation n C k means "n choose k" and can be found using the formula

n C k = (n!)/(k!*(n-k)!)

Answer 14

let me know if that helps at all

Answer 15

Sorry, I had to go eat.
@jim_thompson5910 so it would be:

n C k = (20)/3 * (20 - 3)) = 0.392
(0.392) * (1/6)^(3) * (1 - (1/6))^(20 - 3) =
1.8155E - 03 =
0.0018155

Is that right?

Answer 16

20! and not just 20

Answer 17

the exclamation mark is factorial notation (sorry I should have stated that)

Answer 18

so 20! = 20*19*18...*3*2*1

most calculators have a factorial button so you don't have to type in all that

Answer 19

the same is true for the 3 and the (20-3)

Answer 20

finally someone calls it dice
i am sick to death of "number cubes"

Answer 21

Ohhh...
I don't know how to do that on a calculator o.o

Answer 22

do it in one step in wolfram

Answer 23

what kind of calculator do you have?

Answer 24

Casio fx-9750GII
but I have no clue how to like do anything on it. I know how to do very little on it.

Answer 25

ok let me bring up a users manual on it

the images I'm looking at suggest that this thing is similar to a ti-83 (which I am familiar with)

Answer 26

ok found it

Answer 27

Okay thank you :)
Yeah, it's kinda like a ti but different.

Answer 28

you hit the option button, then you find the PROB submenu

Answer 29

in there you'll see n P r (for permutations) and n C r (for combinations)

we'll use n C r

Answer 30

Okay found it

Answer 31

most likely (based on how the ti83 works), you type it in like this


20 n C r 3

Answer 32

so you type 20 first, bring up the n C r, then you type 3 to finish it off

Answer 33

Alright, I did that and the number 1140 came up

Answer 34

good, that means 20 C 3 = 1140

Answer 35

So

(n C k)*(p)^(k)*(1-p)^(n-k)

(20 C 3)*(1/6)^(3)*(1-1/6)^(20-3)

(1140)*(1/6)^(3)*(1-1/6)^(20-3)

and I'll let you compute the rest

Answer 36

I got 5.277777778
Is that right?

Answer 37

that's way too big, remember probabilities are between 0 and 1

Answer 38

0 meaning 0% (event never happens)

1 meaning 100% (event is guaranteed to happen)

Answer 39

Oh, doesn't that mean I need to divide it by like 100 or something?

Answer 40

not quite, it means that you know what window your answer should be in

but the answer isn't 5%

Answer 41

what are you getting when you compute (1/6)^(3)

Answer 42

4.6296E - 03

Answer 43

so that is also written as 0.0046296

Answer 44

how about (1-1/6)^(20-3)

Answer 45

1

Answer 46

just 1?

Answer 47

it should be some decimal number

Answer 48

so this is probably where the error is happening

Answer 49

probably, because it just came up as 1

Answer 50

what did you type in exactly?

Answer 51

you should have typed in (1-1/6)^(20-3) exactly as you see it

Answer 52

oh and I should mention that 1-1/6 is not 0

it's really 1 - (1/6)

so you could simplify that portion to get 1-1/6 = 5/6

Answer 53

oh I think I did something wrong before because I just did it again and this time I got 0.04507

Answer 54

very good

Answer 55

so

(1140)*(1/6)^(3)*(1-1/6)^(20-3)

turns into

(1140)*(0.0046296)*(0.04507)

Answer 56

of course, you can type in all of (1140)*(1/6)^(3)*(1-1/6)^(20-3) to get the direct answer (but it helps to break it down a bit)

Answer 57

I got 0.23787
Does that look right?

Answer 58

when I typed in (1140)*(1/6)^(3)*(1-1/6)^(20-3), I'm getting 0.23788656613786

So it does look right

Answer 59

So you roughly have a 23.7% chance of rolling exactly 3 sevens

Answer 60

or 23.8% I mean (if you're rounding percentages to 1 decimal place)

Answer 61

Omg yay! Thank you so much!
I have a couple of other problems on this assignment that I could use some help with, if you don't mind could you help? If not, that's okay. But seriously, thank you.

Answer 62

sure, go ahead

Answer 63

Should I put them in a new question or just put them on here?
And thank you!

Answer 64

either way works, a new question is probably best so things don't get too cluttered or laggy

Answer 65

Alright. One sec.

Answer 66

alright