SOLUTION: Is the binomial distribution appropriate for the following situation? Joe buys a ticket in his state�s �Pick 3� lottery game every week; X is the number of times in a year that

Question 1072488: Is the binomial distribution appropriate for the following situation?
Joe buys a ticket in his state�s �Pick 3� lottery game every week; X is the number of times in a year that he wins a prize. Can you show step-by-step how you got the answer Plaese

Yes

No

Cannot determine from information given

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i think it might be appropriate.

the answer to the question should be yes.

here's my explanation.

you have either a win or a lose proposition each week.

let p equal the probability of winning.

let q equal the probability of losing.

since this is a win or lose situation, then q = 1 - p.

let x equal the number of times he wins in a year.

let n equal the number of times he plays.

let p(x) equal the probability of winning x times.

the binomial formula is:

p(x) = p^x * q^(n-x) * c(n,x)

we'll assume that the probability of winning is .05.

this is not accurate, but is ok for example purposes because it's simple enough to be shown clearly.

you get p = .05

since q = 1 - p, you get q = .95

we have n = 52.

x will range from 0 to 52.

the formula for p(x) will become:

p(x) = .05^x * .95^(52-x) * c(52,x)

for example:

when x = 0, this formula becomes:

p(0) = .05^0 * .95^52 * c(52,0)

simplify this to get p(0) = .05^0 * .95^52 * 1 which is equal to .06944.....

c(n,x) is the number of ways you can get sets of x items from n items.

c(n,x) is equal to n! / (x! * (n-x)!)

i used excel to calculate all the probabilities, the sum of which is equal to 1, as it should be.

you can see that p(x = 0) is .0694..... as i had calculated manually.

those calculations are shown below:

$$$

the requirements for the binomial formula can be found in the following reference.

https://people.richland.edu/james/lecture/m170/ch06-bin.html
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