Geometric situations

The geometric distribution is related to the binomial distribution. In the latter, we considered the probability of $x$ "successes" occuring after $n$ trials with a fixed binary probability of success of $p$. In the geometric distribution, we ask a slightly different question: In a series of binomial trials with fixed probability (Bernoulli trials), what is the probability that success will occur on the x^th trial?

The geometric distribution

If the probability of success in a geometric situation is $p$, then the probability of failure — provided that each trial is independent of the others — is $1-p$. If we're asking for the probability that success occurs on the x^th trial, then we need to have failure on the first $x-1$ trials, then the success. Mathametically that is

$$P(X=x) = (1-p)^{x-1} p ,$$

where $X$ is our random variable, and $x$ is the value of $X$ for which we seek the probability.

In our example above that proabability would be

$$ \begin{align} P(X=5) &= (1-p)^{x-1} p \\[5pt] &= (1-0.06)^{5-1}(0.06) \\[5pt] &= 0.94^4 (0.06) = 0.0468 \end{align}$$

or about a 4% chance.

Now let's look at a portion of the distribution by calculating probabilities of finding a pediatrician on the first, second, third, and so on ... trials.

Example 1

Consider flipping a fair coin. Calculate the probabilies that

1. Obtaining the first "heads" on the third flip.
2. Obtaining heads in the first three flips.

Solution: (a) Our random variable is $X$, the number of flips that come up heads. $p$ is the probability of obtaining heads. The probability of obtaining a heads on the third toss is the probability of not tossing heads twice times the probability of tossing heads. It is

$$ \begin{align} P(X=3) &= (1-p)^{3-1}p \\[5pt] &= (1-0.5)^2 0.5^1 \\[5pt] &= 0.125 = 12.5\% \end{align}$$

The probability of tossing heads for the first time on the third flip is 12.5%.

(b) The probability of tossing heads in either toss 1, 2 or 3 is the sum of probabilities,

$$P(X=1) + P(X=2) + P(X=1)$$

That is

$$ \begin{align} P &= 0.5^2 \cdot 0.5 + 0.5^1 \cdot 0.5 + 0.05^0 \cdot 0.5 \\[5pt] &= 0.125+0.250+0.5 = 0.875 \end{align}$$

Example 2

A company with a computer network experiences network failures in about 10% of weeks. Let's say that the leaders of the company would like to know the probability that the company can go 5 weeks or longer without experiencing a failure.

Solution: We'd like to know the probability of failure to have a failure for five weeks. That's just $(1-p)^5$, where $p=0.1$. So the probability is

$$ \begin{align} P(X=5) &= (1-p)^{5-1}(p) \\[5pt] &= (1-0.1)^{5-1}(0.1) \\[5pt] &= 0.0656 \approx 6.6\% \end{align}$$

If we'd wanted to know the probability that a network failure (a "success" in this problem) would occur in the sixth week, we'd caclulate

$$ \begin{align} P(X=6) &= (1-p)^{6-1} p \\[5pt] &= (1-0.1)^5 (0.1) \\[5pt] &= 0.059 \approx 6\% \end{align}$$

There would be about a 6% chance of the failure falling in week 6. To get the proabability that the network would last longer than 5 weeks, we'll have to sum the first five probabilities, then subtract the sum from 1:

$$ \begin{align} P(X\gt 5) &= 1-[P(X=1) + P(X=2) + \\[5pt] &P(X=3) + P(X=4) + P(X=5)] \\[5pt] &= 1-[0.9^4(0.1)+0.9^3(0.1)+0.9^2(0.1) + \\[5pt] & \phantom{000} 0.9^1(0.1) + 0.9^0(0.1)] \\[5pt] &= 0.5905 \approx 60\% \end{align}$$

Practice problems

1. Consider rolling a six-sided die.

   1. Calculate the probability that a five is rolled for the first time on the third roll.
   2. Calculate the probability that a 4 is rolled more than twice before a 5.
   3. Calculate the mean number of rolls before rolling a 5, and its standard deviation.
   Solution

   Answer here

2. Consider rolling a six-sided die.

   1. Calculate the probability that a five is rolled for the first time on the third roll.
   2. Calculate the probability that a 4 is rolled more than twice before a 5.
   3. Calculate the mean number of rolls before rolling a 5, and its standard deviation.
   Solution

   Answer here