How does a geometric model differ from a binomial model?

A binomial model fixes the number of trials and counts the successes, while a geometric model lets the trials run on and counts how many are needed to reach the first success. The four conditions on the trials themselves are the same in both.

A LevelFurther MathsEdexcelFurther Statistics 1FlashcardsDiscrete Probability DistributionsGeometric & Negative Binomial Distributions

Geometric & Negative Binomial Distributions (Edexcel A Level Further Maths: Further Statistics 1): Flashcards

Exam code: 9FM0

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FrontThe Geometric Distribution
Define the geometric distribution.

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Cards in this collection (15)

Define the geometric distribution.
The geometric distribution $Geo (p)$ models the number of trials needed to reach the first success, where $p$ is the constant probability of success in any one trial.
So $X = 8$ means that the first success happened on the eighth trial.
What four conditions must an experiment satisfy for a geometric model?
There must be an indefinite number of successive trials, each independent of the others, each having exactly two outcomes, and a constant probability of success $p$ .
The trials continue until the first success, so the number of them is not fixed in advance.
For $X ~ Geo (p)$ , complete the probability that the first success comes on or before the $x$ th trial:
$P (X \leq x) = 1 - {(______)}^{x}$
The completed result is:
$P (X \leq x) = 1 - {(1 - p)}^{x}$
The first success falls on or before the $x$ th trial exactly when the first $x$ trials are not all failures, and the probability of $x$ failures in a row is ${(1 - p)}^{x}$ .
How does a geometric model differ from a binomial model?
A binomial model fixes the number of trials and counts the successes, while a geometric model lets the trials run on and counts how many are needed to reach the first success.
The four conditions on the trials themselves are the same in both.
True or False?
For a geometric distribution the most likely number of trials is 1.
True.
Every extra trial required is an extra failure first, so the probabilities decrease as the number of trials grows.
That makes $P (X = 1)$ the largest of them, so $X = 1$ is the mode however small $p$ is.
Why is the geometric distribution called geometric?
Because its probabilities $p$ , $p (1 - p)$ , $p {(1 - p)}^{2}$ and so on form a geometric sequence, with first term $p$ and common ratio $1 - p$ .
Each further trial needed multiplies the probability by another factor of $1 - p$ .
True or False?
After five failed trials, a success on the next trial is more likely than it was at the start.
False.
The geometric distribution has no memory, so the probability of success on any single trial is always $p$ , whatever has gone before.
The number of further trials still needed has the same distribution it had at the very beginning.
A geometric distribution has a mean of 500. What is $p$ , and what does that tell you in practice?
Setting $\frac{1}{p} = 500$ gives $p = 0.002$ , so a success is expected about once in every 500 trials.
The mean is the average number of trials needed, so a small $p$ always produces a long average wait.
Define the negative binomial distribution.
The negative binomial distribution models the number of trials needed to reach a fixed number of successes $r$ , where $p$ is the constant probability of success in one trial.
For example, how many rolls of a dice are needed before a six appears for the third time.
What is fixed, and what is counted, in a binomial, a geometric and a negative binomial model?
A binomial fixes the number of trials and counts the successes, while a geometric fixes one success and a negative binomial fixes $r$ successes, both counting the trials needed to get there.
The conditions on the trials themselves are identical in all three, so what is fixed is the only thing that tells them apart.
Why does the negative binomial probability multiply a binomial probability by $p$ ?
Because the $r$ th success has to land on the $x$ th trial itself, which contributes a factor of $p$ .
The first $x - 1$ trials must supply exactly $r - 1$ successes in any order, and that part is an ordinary binomial probability.
For a negative binomial variable needing $r$ successes, complete the range of values $X$ can take:
$X =______, r + 1, r + 2, \dots$
The completed range is:
$X = r, r + 1, r + 2, \dots$
The smallest value is $r$ because $r$ successes cannot be reached in fewer than $r$ trials, and there is no largest value at all.
True or False?
The geometric distribution is a special case of the negative binomial distribution.
True.
Setting $r = 1$ reduces the negative binomial probability function to $p {(1 - p)}^{x - 1}$ , which is exactly the geometric probability function.
Waiting for the first success is simply waiting for $r$ successes when $r$ happens to be 1.
Someone repeats an activity until their $r$ th success. Why might a negative binomial model be criticised here?
Because the trials may not be independent: repeating the activity gives practice, so the person may improve as they go.
That also breaks the constant probability of success, since $p$ would be rising from one trial to the next.
A negative binomial variable has a mean of 10 and needs $r = 4$ successes. What is $p$ ?
Setting $\frac{r}{p} = 10$ with $r = 4$ gives $p = 0.4$ .
When a question supplies the mean or the variance together with one of the two parameters, forming an equation is the route to the other.

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