Question

You roll a pair of dice five times. Find the probability of each of the following outcomes.
a) You get doubles exactly once
b) You get exactly three sums of 7
c) You get at least one sum of 7
d) You get at most one sum of 7

I've been using the formula P(x=k)=(n / k)P^k (1-p)^n-k

For a) I got 0.4019, not sure if that's correct
I'm having trouble with the rest

Answer 1

What distribution is that?

Answer 2

What do you mean?

Answer 3

Ohhhh binomial distribution

Answer 4

Yeah sorry it's the only one I've learned

Answer 5

For the first one I have it set up like this
n=5
success= doubles
p=1/6
x= number of times getting doubles
Then P(x=1)= (5 / 1 )(1/5)^1(1-1/6)^5-1
Where the (5 / 1) isn't supposed to have the / symbol but I've no idea how to show it's five over one without it.

Answer 6

that's little p
I'm not really sure how I'm supposed to do this :/

Answer 7

wait lemme reread my notes here. I may be getting carried away

Answer 8

X= number of times the success should happen.
And since you want doubles only once for the first one, it's one. For b) it should be 3.

Answer 9

That's what I think it means at least.

Answer 10

See here is my issue. What is the probability of getting a double which in this case is our p

Answer 11

How did you figure out what your p is

Answer 12

1-1/6

Answer 13

little p is the amount of times out of two six sided dice doubles could happen, which would be 1/6. Or 6/36

Answer 14

This is how I'm reading my notes, not sure what you're thinking or reading

Answer 15

Ahhhhhh now i seee itttttt

Answer 16

Cool so I'm not wrong for once :p My problem is the next three questions.

Answer 17

Ya but You are wrong in this case cuz p=1/6 so q=5/6

Answer 18

q?

Answer 19

I don't have a q in my formula

Answer 20

P(x=1)= (5 / 1 )(1/5)^1(1-1/6)^5-1

Answer 21

oh gotcha

Answer 22

The real equation should be
P(x=1)= (5 / 1 )(1/6)^1(1-1/6)^5-1

Answer 23

I think I mistyped the equation in the first part, I have it written down correctly on my notes.

Answer 24

which equals .4019 like you stated above

Answer 25

Okk lets go onto the next one then

Answer 26

Okay, so for this one I have P(> or = 3)
So P(X=3)+P(X=4)+P(X=5)

Answer 27

4+3=7
3+4=7
2+5=7
5+2=7
1+6=7
6+1=7
So there 6 ways we can get a sum of 7 when rolling to dies
so p=1/6

Answer 28

Yes, so for P(X=3) I have (5 / 3)(1/6)^3(1-1/6)^5-3

Answer 29

Yup that looks correct

Answer 30

But for (5 / 3) Is that equal to 1 also because it's in (n / n) format?

Answer 31

(5*4*3*2*1)/((3*2*1)(2*1)
20/2=10

Answer 32

I've never been shown that way. Why do you skip 4*3*2*1?

Answer 33

Ok sorry just jumped a couple of steps cuz i thought u followed

Answer 34

I'm sorry it's my first statistics class

Answer 35

\[\frac{5!}{3!(5-3)!}\]
\[\frac{5!}{3!*2!}\]
\[\frac{5*4*3*2*1}{(3*2*1)(2*1)}\]
\[\frac{5*4}{2*1}=\frac{20}{2}=10\]

Answer 36

So that's what you do every time that you have a (5 / 3) or the likes? Not if there is a one involved?

Answer 37

Yup here is the general formula