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Question 175:



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You can use the properties of the binomial distribution to answer these questions. You would use the binomial cdf -cumulative distribution function and the pmf probability mass function, depending on if the question is asking you to provide the probability of a cumulative range (cdf) or just the probability of  exactly one value (pmf). The formula for the Binomial pmf is:

 n  p x (1-p)n-x

n= number of trials

p= probability of selecting

x = number of successes

n-x = number of failure

1-p = probability of failure

The n over the x means n choose x and is the number of ways to get x successes out of n trials.


To answer your questions:

  1. The probability that all 4 of 4 of the 1st calls will all be Republican is asking for just one value, not the whole range. So you plug and chug or use Excel's =BINOMDIST(4,4,0.3,FALSE) which provides with a .81% chance of that occurring (the same as multiplication rule in this case .3*.3*.3*.3 = .081.)
  2. Exactly 2 Republicans is just like the 1st, except you have more ways of doing it, but we still use the pmf =BINOMDIST(2,4,0.3,FALSE) or abut 26.5% chance of that occurring.
  3. For no independents, it's also the pmf, because its asking again for an exact number. The only difference here is the probability of selecting an independent is .11 so the formula in excel would read  =BINOMDIST(0,4,0.11,FALSE) or about a 62.7% chance of that happening.
  4. For all independents, it would be an unusual event and basically the opposite of the 1st question (.11*.11*.11*11) or using the Binomial pmf =BINOMDIST(4,4,0.11,FALSE) provides a .015% chance of this occurring.

Stop for a second, and be sure it intuitively makes sense. Since there aren't a lot of independents (11 in 100) compared to Democrats (35 in 100) we should not expect to be reaching a lot of independents in a sample of 4. We should expect the probability of selecting all independents to be much lower than selecting 0 out of 4, which is what we're seeing.

At least one independent : Finally this one has you use the cdf, which is just adding up (accumulating) the values from the pmf in choosing 1, 2, 3 or 4 independents. For choosing 1 its =BINOMDIST(1,4,0.11,FALSE) = .31. Rinse and repeat for 2, 3 and 4. Then add them all up. You should get around a 37% chance of selecting at least one independent in 4 calls.

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