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# Probability

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?

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Kay G. | ~20 Years Accounting Tutoring Experience~20 Years Accounting Tutoring Experience
4.9 4.9 (32 lesson ratings) (32)
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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) (px)[(1-p)(n-x)]

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