Discrete Probability Functions#

Certain probability scenarios arise so often that we create and utilize functions for them. These functions are called probability density functions and are often abbreviated as pdf (or pmf for probability mass fucntions). We will learn to create the pdf for four specific distributions:

  1. Binomial

  2. Geometric

  3. Hypergeometric

  4. Negative Binomial

Binomial Distributions#

Binomial, in a probability sense, indicates situations were only two possiblities exist. In probability problem-solving, we often refer to the outcomes as successes or failures.

The idea of “success” has nothing to do with whether the outcome is preferable, only that it is the outcome whose probability we are asked to calculate. For example, when calculating the probability of rain on a proposed date for an outdoor wedding, we might term the outcome RAIN as a success while solving the problem. The bride would likely disaggree with this characterization.

Example 10

Betting on Red or Black in roulette means that you either you win, or you lose all the money you bet. Winning means that your bet is doubled (nearly) and the amount is returned to you.

Chelsea plays 5 rounds of roulette. Find the probability that she wins 3 or more of her 5 roulette plays given that she plays either Red or Black every time.

Solution. Assume the outcomes of the spins are perfectly random amongst the landing slots labeled 0, 00, 1, 2, … ,35, 36. Thus, every bet on Red or Black has the same probability of success. For 18 of the 38 total slots, Chelsea wins. Thus:

\[P = \frac{18}{38} = \frac{9}{19} \approx 0.474\]

The payout for a win is \(2:1\) which would be fair odds if the probability of success were exactly \(\frac{1}{2}\). First, let’s consider the probability of Chelsea playing 5 spins of the wheel and winning exactly 3 times. Suppose the pattern of outcomes was 3 wins first followed by 2 losses:

_images/roulette.png

We need to calculate the probability of win then win then win and then a loss and a loss. All are independent trials:

\[P(WWWLL) = \left(\frac{9}{19}\right)\left(\frac{9}{19}\right)\left(\frac{9}{19}\right)\left(\frac{10}{19}\right)\left(\frac{10}{19}\right)\hspace{3mm}\implies\]
\[P=\left(\frac{9}{19}\right)^3\left(\frac{10}{19}\right)^2\]

Notice that the probability would be identical if the order of the wins and losses were changed. In fact, there are \(\binom{5}{3}=10\) total patterns:

  • W W L W L

  • W L W W L

  • W L W L W

  • L W W W L

\(\vdots\)

Let the simple random variable (SRV) \(n\) denote the number of wins out five attempts, and we have:

\[P(n=3) = \binom{5}{3}\left(\frac{9}{19} \right)^3\left(\frac{10}{19} \right)^2\]

Binomial Distribution Formula#

Given the following values:

  • \(n\) is the number of total trials,

  • \(k\) is the number of successes,

  • \(p\) is the probability of success, and

  • \(q=1-p\) indicates the probability of failure.

The probability of \(k\) successes in \(n\) binmoial trials is given by:

\[P(k)=\binom{n}{k}p^kq^{10-k}\]

In the binomial distribution, we might asked the probability of at least \(t\) successes in \(n\) trials where \(1\leq t <n\). Then we must sum the exact probabilities up as follows:

\[P(k\geq t) = \sum_{k=t}^n\binom{n}{k}p^kq^{(n-k)}\]

Geometric Distribution#

A fair, 6-sided die is rolled repeatedly until a “3” appears. If \(n\) is the SRV indicating the number of trials required to see a 3:

  • \(P(n=1) = \frac{1}{6}\)

  • \(P(n=2) =\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)\)

  • \(P(n=3)=\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)\)

How does the pattern continue?

The pdf#

Given the following values:

  • \(n\) is the index of the first success,

  • \(p\) is the probability of success, and

  • \(q=1-p\) indicates the probability of failure.

The probability of the first success occuring on the \(n^{th}\) trial is given by:

\[P(n) = (q)^{n-1}p\]

Hypergeometric Distribution#

A deck of cards contains 20 cards: 6 red cards and 14 black cards. 5 cards are drawn randomly without replacement. What is the probability that exactly 4 red cards are drawn?

\[P=\frac{\binom{6}{4}\binom{14}{1}}{\binom{20}{5}}\]

The pdf#

Given the following values:

  • \(N\) total objects

  • \(K\) objects which count as successes

  • \(n\) total draws attempted

  • \(k\) successful draws

The probability of \(k\) successful draws out of \(n\) total draws:

\[P(k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\]

Negative Binomial#

Given the following values:

  • \(n\) indicates the number of trials.

  • \(k\) indicates the number of successes needed for overall success.

  • \(p\) indicates the probability of success.

  • \(q=1-p\) indicates the probability of failure.

  • \(X\) represents the number of the last well drilled to achieve overall success.

The pdf#

The closed form pdf for a negative binomial distribution is as follows:

\[P(X=x) = \binom{n-1}{k-1}p^kq^{n-k}\]

Example

An oil company conducts a geological study that indicates that exploratory oil wells in a certain basin should have a 20% chance of striking oil. Due to finances, they need 4 of the first 10 wells to strike oil. What is the probability that the \(4^{th}\) successful strike occurs on the \(10^{th}\) well drilled?

Notice the pattern requires the \(10^{th}\) well to be a success:

* * * * * * * * * S
Thus, we have 9 wells amongst which we distribute 3 successes and 6 failures. The probability is therefore:
\[P(X=10) = \binom{9}{3}\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^6\]

Note that our intuitive approach to patterns and counting led to the same place as using the formula would have. This should provide strong encouragement to understand the patterns and counting so that, on a quiz or test, you may be able to use logic and problem-solving to confirm the formula is providing you with the correct answer.