Probability

Question

Can you walk me through this please? Thank you!

 

NFL. On any given Sunday, any team could beat any other team. If we assume every week a team has 50% chance of winning, what is the probability the team will have at least one win?

 

For a particular brand of generators, 20% of the ones on the market are defective. If a company buys 10, what is the probability that none are defective?

 

 

 

Answer 1

These are binomials.

 

I'm not getting how you can do the first one without a number of weeks.  Maybe I'm just not thinking of something.

 

So I'll use the second problem.

 

There's a binomial equation:

P(n,x) = C(n,x) (p^x)[(1-p)^(n-x)]

 

Sometimes (1-p) is called q, so it might look less icky to write it C(n,x)(p^x)(q^(n-x))

 

OK, so it's still icky looking. :-)  (There's a bit of a pattern to it if you'll think about it.)

 

Let's insert your numbers:

C(10,0)(.20⁰)(.80¹⁰)

 

Notice the .20 is the proportion you were given that are defective, 20%, or .20.  So p = .20.  Then you also want the probability of not defective, which is 1 - p, or 1 - .20 = .80.  If .20 are defective then .80 are not.

 

The n is your sample, 10.  You bought 10.  The x is the number you're interested in, which in this case is none, i.e. 0.  So you're looking for 0 out of 10 to be defective.

 

You start with the combination, 0 out of 10.

 

Notice the exponents are the 0 for the .20.  And the .80 (not defective) is the difference between n and x, or the number not defective.  (If 0 are defective, then 10 would be not-defective.)

 

So that's how I plugged into the equation the way I did.

 

Are you able to solve that equation from there?

 

Just for example's sake, what if they wanted to know the probability of 2 being defective if you purchased 15.  This also means 13 would not be defective:

 

C(15,2)(.20²)(.80¹³)

 

Just wanted you to see one that doesn't have that 0 in it.