Binomial Distribution

Let's now apply this to probability. Suppose you throw a die 10 times as in problem 1.3.6, and this time you want to know a lot more. Hey, we're scientists here, insanely detailed questions really float our boats. So what we'll ask are things like: what's the probability of getting exactly 3 2's when we toss the die 10 times. Suppose if you get a 2, you denote this by "a" and if you don't you denote it by "b". Then a possible outcome of the sort we're looking for is babbbabbab, or bbaabbbabb. Sounds like an insane sheep. But how many such possibilities are there? Well this is exactly the same counting argument that went into the Binomial Theorem. The answer is $\left(\!\!\! \begin{array}{c} 10 \\ 3 \end{array} \!\!\!\right) = 10!/(7!3!) = 10~9~8/6 = 120$ possibilities.

What's the probability of getting a single one of these outcomes, e.g. babbbabbab. Well the probably of getting a 2 . And getting something else is . So because these events are independent,

$\begin{displaymath} P(babbbabbab) = P(b)P(a)P(b)P(b)P(b)P(a)P(b)P(b)P(a)P(b) = (1/6)^3 (5/6)^7. \end{displaymath}$

(1.8)

So the total probability of getting exactly 3 2's is $\left(\!\!\! \begin{array}{c} 10 \\ 3 \end{array} \!\!\!\right) (1/6)^3 (5/6)^7 = .155$

We can now generalize this to any situation where you have independent trials. The probably of getting is . Then the probability of getting exactly 's leads the binomial distribution formula:

$\begin{displaymath} \left(\!\!\! \begin{array}{c} n \\ m \end{array} \!\!\!\right) (p)^m (1-p)^{n-m} \end{displaymath}$

(1.9)

and is exactly the same as an individual term in the binomial expansion.

Keywords

binomial distribution formula, binomial expansion, combination, pascal's triangle feature, pascal's triangle worksheet, probability, problems.

Subsections

Problems

josh 2010-10-20