statistics help needed

N

natenatmom

New member
Joined
Dec 17, 2005
Messages
24
I am totally confused on how to do this. The questions is - when rolling a die 4 times, what is the probability that you will roll a 6 no more than twice. Each roll is independent which means you have a .166 chance of rolling a 6 each time. But I just don't know where to go from there. Can anyone help?
 
Rolling a die four times, the probability of getting a six three times is:
P(x=3)=4(1/6)<sup>3</sup>(5/6).
The probability of four sixes is P(x=4)= (1/6)<sup>4</sup>.
Now the probability of NO MORE THAN TWO SIXES is the opposite of that:
P(x≤2)=1−[4(1/6)<sup>3</sup>(5/6)+ (1/6)<sup>4</sup>]
 
One way is the binomial thing.

Expand \(\displaystyle (p+q)^{4}\)

\(\displaystyle p^{4}+4p^{3}q+6p^{2}q^{2}+4pq^{3}+q^{4}\)

The probability of rolling a six is 1/6.

Now, since it says, "no more than twice", use \(\displaystyle q^{4}+4pq^{3}+6p^{2}q^{2}\)

Sub in 1/6 for p and 5/6 for q.
 
Thank you all so much! That really helps. Now I've got a similar problem so tell me if I'm doing this right... There's a .13 probability that a police officer issues no tickets on a given day. There's a .14 probability of 1 ticket, .43 for 2 tickets, .20 for 3 tickets, and .10 for 4 tickets. What is the probability that over the next 5 days he will issue no tickets on exactly one day? So would I have P(x=0) = 5(.13^1)(.77^4) since we have 1 day with the probability for no tickets and 4 days with the probability of any number of tickets except 0? I have .23 as my answer.
 
I can help you with all these problems... just email me and let me know!

Edit: Link Removed for Personal Security
 
natenatmom said:
Thank you all so much! That really helps. Now I've got a similar problem so tell me if I'm doing this right... There's a .13 probability that a police officer issues no tickets on a given day. There's a .14 probability of 1 ticket, .43 for 2 tickets, .20 for 3 tickets, and .10 for 4 tickets. What is the probability that over the next 5 days he will issue no tickets on exactly one day? So would I have P(x=0) = 5(.13^1)(.77^4) since we have 1 day with the probability for no tickets and 4 days with the probability of any number of tickets except 0? I have .23 as my answer.
Click to expand...

I believe that should be .87, 1-.13=.87

\(\displaystyle 5(.13)(.87)^{4}=37.2%\)
 
ldoron said:
I can help you with all these problems... just email me and let me know! [ad links deleted]
Click to expand...
How about helping people, instead of just posting ads for your business? :roll:

Eliz.
 
Ok guys, the teacher says that answer is wrong (the one about the tickets) but won't explain why. He just says to look at the book and redo it. I have no idea! Am I using the wrong formula?
 
You had the formula almost right. Over the next 5 days, the probability that exactly one day (out of 5) has zero tickets issued and 4 days have more than zero tickets is \(\displaystyle 5(.13)^4(.87)\). This is a straight binomial calculation, just remember that the probabilities must add to 1.
 
Ok, so I had the wrong one to the 4th power. Man, this is confusing! So why is .13 to the 4th when that's the probability for 0 tickets which we only want on 1 day?
 
Oops, I read it backwards. Galactus has the correct answer above: \(\displaystyle 5(.13)(.87)^4\), since you want exactly one out of 5 possible days with zero tickets, and four with at least 1 ticket. What I calculated is the probability of zero tickets for four out of 5 days.
 
Well what I turned in was what Galactus had and the teacher said it's wrong. I don't understand what's wrong with it.
 
There is nothing wrong with it. That is the solution for the problem you posed. If the teacher got the answer from the back of a book, the answer could be wrong. I teach an elementary statistics course for college students, and they frequently find "back of the book" answers that are wrong.
 
Thanks, Royhaas. The teacher emailed and said to look at his notes which are just about the worst set of notes I've ever seen. He just writes down a bunch of sample problems, doesn't explain what he's doing or when to use which formulas. The only thing I can find that might be what he wants is this...

p(x) = n! p^x q^n-x
X!(n-x)!


That X!(n-x)! should be right under the n! Would that work for this problem? Here's the problem again... What is the probability that over the next 5 days no parking tickets will be issued on exactly one day? The probabilities are f(0) = .13, f(1) = .14, f(2) = .43, f(3) = .20, f(4) = .10
 
What you have written is the formula for binomial probabilities. However, it's the same as what has already been written, with \(\displaystyle n=5,p=.13,q=.87,x=1\).
 
The detailed wording of the question leads me to think that there is more to it than appeared. Is this the EXACT question? “What is the probability that over the next 5 days he will issue no tickets on exactly one day?"
 
You must log in or register to reply here.
Top