More Probability

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studybrat

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Oct 18, 2005
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Okay this one confused me a bit:

A new lottery has just been started. The lottery of 10 numbers (frm 0-9) and the participant must choose 4 numbers from those 10. What is the prob. that a participant who buys one lottery ticket will come up with the same four numbers as those chosen at random by the lottery.

I started taking 4 numbers away from 10, but they can be used over again, right? So, would I mult. 10 * 4 ??


Also, this hot dog question has me too. I'm not sure where to start on these types of problems.

Elly's hot dog emporium is famous for its chilidogs. Elly's latest sales indicate that 30% of the customers ordering her chilidogs order it with peppers. Suppose 18 customers are selected at random. What is the prob. that exactly ten customers will ask for hot pepper?

I just did a Poisson distribution prob. and thought that's how I was supposed to set this problem up, but I got stuck.

I took .30 * 18 = 5.4 , but when I went to look at the chart the book has in the back, I wasn't sure what to do with the .4 part. The books examples in Poisson distrib. use only whole numbers, but then again I may be going about this all wrong.

Lottery and Hot Dog Help would be greatly appreciated!
 
A new lottery has just been started. The lottery of 10 numbers (frm 0-9) and the participant must choose 4 numbers from those 10. What is the prob. that a participant who buys one lottery ticket will come up with the same four numbers as those chosen at random by the lottery.
Click to expand...

First note that lotteries tend to withdraw numbers without replacement.

To begin there are 10 numbers, four of which are winners. So P(1st a winner) = 4/10.

There are now 9 numbers remaining, three of which are winners. P(2nd a winner) = 3/9

And so on for the thid and fourth selections.

So we have P(four winners) = 4/10 * 3/9 * 2/8 * 1/7 = 1/210.
 
Elly's hot dog emporium is famous for its chilidogs. Elly's latest sales indicate that 30% of the customers ordering her chilidogs order it with peppers. Suppose 18 customers are selected at random. What is the prob. that exactly ten customers will ask for hot pepper?

I just did a Poisson distribution prob. and thought that's how I was supposed to set this problem up, but I got stuck.

I took .30 * 18 = 5.4 , but when I went to look at the chart the book has in the back, I wasn't sure what to do with the .4 part. The books examples in Poisson distrib. use only whole numbers, but then again I may be going about this all wrong.
Click to expand...

A poisson distribution applies to rare events. You tend to be given a rate at which an event occurs (e.g on average, 1 person gets hit by a flying pepper every month).

Here we are given the probability of a customer having peppers. A customer either has peppers or does not. Thus a binomial distribution is in order.

pi = ?, n = ?, x = ?
 
Absolutely correct! Your binomial distribution tables probably won't have n=18 in them, so the binomial formula is in order.
 
Binomial Formula

I'm getting really large #'s; am I doing this right?

P(10) 18! / 1! (18-10)! =

18! / 1! (8)!

That means it's:

18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2
___________________________________________

(1) 8*7*6*5*4*3*2*1


Even with cancelling out the bottom numbers, it's still large:

18*17*16*15*14*13*12*11*10*9

My formula looks like this:

1.587890304^11 (.30)^10 (1-.30)^8

Can this be right?
 
It looks as though you are trying to find 18C10, which is the first part of the binomial distribution formula.

In which case:
18C10 = 18! /[ 10! (18-10)! ]
Click to expand...

If you have a scientific calculator, you can just enter <18> < nCr> <10>.

Once you have that, plug it into

P(x=10) = nCx * pi^x * (1-pi)^(n-x)
 
Sheesh, All For Hot Dogs

Okay, I have come up with an answer of .0148954741

Is that what you come?

I took:

1764322560
__________

40320 = 43758

Then, I plugged it in:

43758(.30)^10(1-.30)^8
 
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