Calculating Binomial Probabilities

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Key Questions

  • Answer:

    See below on my thoughts:

    Explanation:

    The general form for a binomial probability is:

    #sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))#

    The question is Why do we need that first term, the combination term?

    Let's work an example and then it'll come clear.

    Let's look at the binomial probability of flipping a coin 3 times. Let's set getting heads to be #p# and of not getting heads #~p# (both #=1/2)#.

    When we go through the summation process, the 4 terms of the summation will equal 1 (in essence, we are finding all the possible outcomes and so the probability of all the outcomes summed up is 1):

    #sum_(k=0)^(3)=color(red)(C_(3,0)(1/2)^0((1/2)^(3)))+color(blue)(C_(3,1)(1/2)^1((1/2)^(2)))+C_(3,2)(1/2)^2((1/2)^(1))+C_(3,3)(1/2)^3((1/2)^(0))#

    So let's talk about the red term and the blue term.

    The red term describes the results of getting 3 tails. There is only 1 way for that to be achieved, and so we have a combination that equals 1.

    Note that the last term, the one describing getting all heads, also has a combination that equals 1 because again there is only one way to achieve it.

    The blue term describes the results of getting 2 tails and 1 head. There are 3 ways that can happen: TTH, THT, HTT. And so we have a combination that equals 3.

    Note that the third term describes getting 1 tails and 2 heads and again there are 3 ways to achieve that and so the combination equals 3.

    In fact, in any binomial distribution, we have to find the probability of a single kind of event, such as the probability of achieving 2 heads and 1 tails, and then multiplying it by the number of ways it can be achieved. Since we don't care about the order in which the results are achieved, we use a combination formula (and not, say, a permutation formula).

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