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Geometric Distribution

Geometric Distribution

When I was little, my mom used to take me with her to the groceries store every Sunday. There I saw a beautiful stuffed bear which I wanted with all my heart. The problem is that it was inside a claw machine, and my mom would give me just one chance to get it every time we went to the store.

At first, I was confident enough because the game looked easy for me, and every time I would be able to grab the bear with no problem at all. The thing is that every single time the claw just went loose and dropped my bear! After some weeks, tears burst out of my eyes when I was finally able to claim my prize, which I still treasure in my bedroom.

You might be wondering how this relates to probability distributions. Turns out that the claw machines are built so the prize is rarely obtained, no matter how precise your inputs are. In my stuffed bear predicament, I was doing a trial every Sunday until I got a success. In this context, the number of trials that I made until I got my success is represented by a random variable with geometric distribution.

Geometric Distribution Definition

When talking about probability distributions you need to have a clear grasp of which is the random variable you are dealing with. Just like in the stuffed bear example, where I was counting how many times I had to play the claw machine, in a geometric distribution you count how many trials you perform until you obtain a success. It is assumed that each trial is a Bernoulli trial.

Remember that a Bernoulli trial only has two outcomes: success or failure.

It is time to properly define the geometric distribution.

The geometric distribution, also known as the geometric probability model, is a discrete probability distribution where the random variable \( X\) counts the number of trials performed until a success is obtained.

Since the least amount of trials required to obtain a success is \(1\), then the random variable \(X\) can take the values

\[ X=1,2,3, \dots\]

The geometric distribution has only one parameter, which is the probability \(p\) of success. A geometric distribution with probability \(p\) is usually denoted

\[\text{Geom}(p),\]

or sometimes it is written as

\[ G(p).\]

In my stuffed bear example, the random variable \(X\) counted how many times I performed the trial of playing the claw machine until I got my hands on the bear. The probability of success, \(p\), was not known to my person, but in most cases you will be given this value.

A probability distribution needs to satisfy the following requirements in order to fit a geometric model:

  1. There are only two possible outcomes for each trial, success or failure. For example, the first trial could either be a success or a failure, just like all subsequent trials. It is worth noting that the experiment stops once you get a success.

  1. The trials are independent of each other. For example, if the second trial is a failure this will not affect the next trial, or any subsequent trials, in any way.

  1. The success probability remains unchanged trial after trial. This means the probability of success for the first trial is the same for all subsequent trials. For example, if \(p = 0.4\) then the probability of success of the first trial is \(0.4\), the probability of success of the second trial is \(0.4\) as well, and so on.

It is worth noting that if \(p<1\), it is in theory possible that you never obtain success even if you do a large amount of trials. This is easier to picture if \(p\) is a very small number.

Suppose you buy a lottery ticket every month. The chances of actually winning the lottery are astronomically small, so it is most likely that you will never win the prize. How sad!

Formulas Used in the Geometric Distribution

Usually, when you are given a geometric distribution you will be also given some formulas to find certain values of interest.

Probability Mass Function

Since in a geometric distribution you are counting how many trials you take until getting a success, a natural question that arises is: What is the probability of getting the success in exactly \( x\) trials? This can be found by noting that, if you underwent \(x\) trials until you got the success, then you had \(x-1\) failures, so

\[ P(X=x) = (1-p)^{x-1}p,\]

where \(p\) is the probability of success, and \(1-p\) is the probability of failure. You might also find this formula written as

\[ P(X=x) = q^{x-1}p,\]

where \(q=1-p\).

Geometric Distribution six points of the probability mass function where P(x) decreases as x increases StudySmarterFigure 1. Graph of the probability mass function of the geometric distribution

Cumulative Distribution Function

You can find a more realistic approach to an experiment by looking at the cumulative distribution function of the geometric distribution, which tells you the probability of getting success in \(x\) trials or less. For the geometric distribution, this is given by

\[P(X\leq k) = 1-(1-p)^k.\]

Think of the stuffed bear example. Suppose you go to the claw machine with five spare quarters, the cumulative distribution function will tell you the probability of having at least one success with those five quarters, that is

\[ P(X \leq 5) = 1-(1-p)^5.\]

Geometric Distribution six points of the cumulative distribution function where P(x) increases as x increases StudySmarterFigure 2. Graph of the cumulative distribution function of the geometric distribution

Expected Value

The expected value (also known as mean) of the geometric distribution gives you a rough estimate of how many trials you will need to do until you get a success, and it is given by

\[ \mu = \frac{1}{p}.\]

Standard Deviation

The standard deviation, in general, gives you insight on how a variable tends to stay around the expected value. A geometric distribution with a small standard deviation expects the number of trials to be close to the mean. It is given by

\[\sigma = \sqrt{\frac{1-p}{p^2}}.\]

Variance of the Geometric Distribution

Sometimes you will be asked to find the variance of an experiment modeled by a geometric distribution. To make things simple, since the standard deviation is the square root of the variance, you can obtain the variance by squaring the standard deviation. That is if the standard deviation is given by

\[ \sigma = \sqrt{\frac{1-p}{p^2}}\]

then, the variance is given by

\[ \sigma^2 = \frac{1-p}{p^2}.\]

The Geometric Distribution vs. the Exponential Distribution

Because the graph of a geometric distribution looks like a decreasing exponential function, you might associate a geometric distribution with an exponential distribution.

Geometric Distribution an exponential function along with six points of the geometric distribution where both functions are decreasing  StudySmarterFigure 3. An exponential function that passes through the points of the graph of the probability mass function of a geometric distribution

The exponential distribution is quite similar to the geometric distribution in the sense that it models the time-lapse of an experiment until success is obtained. However, because time is considered a continuous quantity, the exponential distribution is a continuous probability distribution, while the geometric distribution is discrete.

Geometric Distribution Examples

Here you can solve some problems that can be modeled using the geometric distribution.

A patient suffers kidney failure and requires a transplant from a suitable donor. The probability that a random donor will match this patient’s requirements is \(0.2\).

  1. Suppose that no donor matches the patient's requirements until a fifth donor comes in. What is the probability of this scenario?
  2. Find the probability of the patient requiring \(10\) or fewer donors until a match is found.
  3. What is the expected number of donors required to get a match?
  4. Find the standard deviation of this scenario.

Solution:

  1. Whenever you need to find the probability that the experiment requires an exact number of trials to succeed, you should start by writing its probability mass function. In this case, since \(p=0.2\) then\[ \begin{align} P(X=x) &= (1-p)^{x-1}p \\ &= (1-0.2)^{x-1}(0.2) \\ &= (0.8)^{x-1}(0.2). \end{align}\]Now, you can evaluate the above function when \(x=5\), giving you\[ \begin{align} P(X=5) &= (0.8)^{5-1}(0.2) \\ &= (0.8)^4(0.2) \\ &= 0.08192, \end{align}\]which means that the probability that this scenario happens is \( 8.192 \%\).
  2. This time you will need the cumulative distribution function, which in this case is given by\[ P(X\leq k) = 1-(1-p)^k.\]Since you are looking for the case where ten or fewer donors are required, you need to plug in \(k=10\) into the above formula (and \(p=0.2\) as well), which will give you\[ \begin{align} P(X\leq 10) &= 1-(1-0.2)^{10} \\ &= 1-(0.8)^{10} \\ &= 0.892625, \end{align}\]so the probability of finding a suitable kidney from ten random donors is of about \(89.26 \%\).
  3. This is a rather straightforward task. For the expected number of donors you should use the formula for the expected value, so\[ \mu = \frac{1}{p}.\]By substituting \(p=0.2\) you will obtain\[ \begin{align} \mu &= \frac{1}{0.2} \\ &=5. \end{align}\]
  4. Finally, you can find the standard deviation by using the formula\[ \sigma = \sqrt{\frac{1-p}{p^2}}.\]Substituting \(p=0.2\) will give you\[ \begin{align} \sigma &= \sqrt{ \frac{1-0.2}{0.2^2} } \\ &= \sqrt{20} \\ &= 4.472133. \end{align}\]

You are likely to find the geometric distribution when playing board games!

Suppose you roll a fair dice until you get a three as a result.

  1. What is the probability that you don't roll a three until your fourth roll?
  2. Find the probability of getting the three you need in less than \(10\) rolls.
  3. What is the expected number of rolls required to get your desired outcome?
  4. Find the variance of this experiment.

Solution:

  1. In this case you need to find the probability of getting the success. Since you are using a fair dice, the odds of getting either number are all equal, so \[ p = \frac{1}{6}\]for obtaining any specific number, which includes getting three as a result. Now that you know \(p\), you can write the probability mass function for this geometric experiment, that is\[ \begin{align} P(X=x) &= (1-p)^{x-1}p \\ &= \left( 1- \frac{1}{6} \right)^{x-1} \left( \frac{1}{6} \right) \\ &= \left( \frac{5}{6} \right) ^{x-1} \left( \frac{1}{6} \right). \end{align} \] Finally, evaluate the above expression when \(x=4\), obtaining\[ \begin{align} P(X=4) &= \left( \frac{5}{6} \right) ^{4-1} \left(\frac{1}{6} \right) \\&= 0.0964506. \end{align}\]This means the probability that you don't get a three until your fourth roll is \( 9.645 \% \).
  2. For this case you will need the cumulative distribution function, which in this case is\[ P(X\leq k)=1-(1-p)^k.\]Here you are asked to find the probability of getting the success in less than \(10\) rolls, which means \(9\) rolls or less, so \( k=9\). Knowing this, you can substitute \(k\) and \(p\) to find the requested probability, so\[ \begin{align} P(X\leq 9) &= 1-\left(1-\frac{1}{6} \right)^9 \\ &= 1-\left(\frac{5}{6}\right)^9 \\ &= 0.806193. \end{align} \]So the probability of getting your desired result in less than \(10\) rolls is \( 80.6193 \% \).
  3. You can use the formula\[ \mu = \frac{1}{p}\]to find the expected value, so\[ \mu = \frac{1}{\frac{1}{6}}, \] which you can simplify with the properties of fractions, giving you\[ \mu = 6.\]
  4. This time you can use the variance formula,\[ \sigma^2 = \frac{1-p}{p^2},\]so\[ \begin{align} \sigma^2 &= \frac{1-\frac{1}{6}}{\left(\frac{1}{6}\right)^2} \\ &=\frac{\frac{5}{6}}{\frac{1}{36}} \\ &= 30. \end{align} \]

Let's assign a number to the probability of succeeding in the claw machine game.

Suppose that the probability of winning an item from a claw machine is \( 0.05\).

  1. What is the probability of winning an item on your first try?
  2. What is the probability of winning an item in less than \(20\) tries?
  3. Suppose you need to use a quarter for each try. What is the expected amount of money spent for getting a prize?

Solution:

  1. This is a tricky question! You can try building the probability mass function and using \(x=1\), but you are already told that the probability of winning an item from the claw machine is \(0.05\), or \( 5\%\), so this is the answer.
  2. As usual, build the cumulative distribution function, so\[ P(X\leq k) = 1-(1-p)^k.\]You need to find the probability of wining an item in less than \(20\) tries, which means \(19\) or less tries. So \(k=19\). Knowing this, evaluate the cumulative distribution function, that is\[ \begin{align} P(X\leq k) &= 1-(1-0.05)^{19} \\ &= 1-(0.95)^{19} \\&=0.622646.\end{align}\] So, the probability of winning a prize in less than \(20\) tries is \(62.2646\%\).
  3. Whenever you are asked about expectations, you should begin by finding the expected value. In this case this means that\[ \begin{align} \mu &= \frac{1}{\mu} \\ &= \frac{1}{0.05} \\ &= 20. \end{align}\]This means that you can expect to play the claw machine about \(20\) times. Since each time you play costs you a quarter, you need \(20\) quarters, so\[20(0.25) = 5\]means that you can expect to spend \($5\) on the claw machine.

Geometric Distribution - Key takeaways

  • The geometric distribution, also known as the geometric probability model, is a discrete probability distribution where the random variable \( X\) counts the number of trials performed until a success is obtained.
    • Since the least amount of trials required to obtain a success is \(1\), then the random variable \(X\) can take the values \( X=1,2,3, \dots\).

  • In order to model a situation using a geometric distribution, you need to make some assumptions: 1. There are only two possible outcomes of a trial, a success or a failure. 2. The trials are independent of each other. 3. The success probability remains unchanged trial after trial.

  • The formulas used in geometric distributions are the following:

    • The probability mass function is given by\[ P(X=x) = (1-p)^{x-1}p.\]

    • The cumulative distribution function is\[ P(X \leq k) = 1-(1-p)^k.\]

    • The expected value can be found as\[ \mu = \frac{1}{p}.\]

    • The standard deviation is\[ \sigma = \sqrt{\frac{1-p}{p^2}}.\]

  • The exponential distribution is similar to the geometric distribution in the sense that both describe situations in which you are looking for the first success of a trial. However, the exponential distribution is a continuous distribution, while the geometric distribution is a discrete distribution.

Frequently Asked Questions about Geometric Distribution

The geometric distribution is a discrete probability distribution where the random variable counts the number of trials performed until a success is obtained.

In order to model a situation using a geometric distribution you need it to meet the following conditions:

  • There are only two possible outcomes for each trial: Success, or failure.
  • The trials are independent of each other.
  • The success probability remains unchanged trial after trial.

You can find a geometric distribution whenever you need to count the number of tries until you get a success in an experiment. 

The geometric distribution is used when you need to count the number of tries until you get a success in an experiment. For example, flipping a coin until you get heads, rolling a fair dice until you get a certain number, buying a lottery ticket until you get a prize.

The mean of the geometric distribution is given by 1/p, where p is the probability of success. The variance is given by (1-p)/p^2.

Final Geometric Distribution Quiz

Question

The geometric distribution is a(n) ____ probability distribution.

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Answer

discrete.

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Question

The exponential distribution is a(n) ____ probability distribution.

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Answer

continuous.

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Question

Suppose you flip a coin five times and count the number of heads obtained. Can this scenario be modeled by a geometric distribution?

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Answer

No.

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Question

Suppose you flip a coin until you get a head. Can this scenario be modeled by a geometric distribution?

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Answer

Yes.

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Question

Suppose you roll a fair dice until you obtain a 5. Can this scenario be modeled by a geometric distribution?

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Answer

Yes.

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Question

Suppose you simultaneously roll two fair dies until the sum of the faces equals 11. Can this scenario be modeled by a geometric distribution?

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Answer

Yes.

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Question

True/False: In a geometric distribution there are only two possible outcomes for each trial.

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Answer

True.

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Question

True/False: In a geometric distribution the trials are independent of each other.

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Answer

True.

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Question

True/False: In a geometric distribution the success probability of each trial changes as you perform more trials.

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Answer

False.

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Question

Which of the following expressions is the probability mass function of the geometric distribution?

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Answer

\( P(X=x) =(1-p)^{x-1}p\).

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Question

Which of the following expressions is the cumulative distribution function of the geometric distribution?

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Answer

\( P(X\leq k) = 1-(1-p)^k\).

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Question

Which of the following expressions gives you the mean of the geometric distribution?

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Answer

\( \mu = \frac{1}{p}\).

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Question

Which of the following expressions gives you the standard deviation of the geometric distribution?

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Answer

\( \sigma = \sqrt{\frac{1-p}{p^2}}\).

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Question

You want to use the cumulative distribution function
\[ P(X \leq k) = 1-(1-p)^k \]
to find the probability that an experiment requires less than \(20\) trials to succeed. What is the value of \(k\)?

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Answer

\(19\).

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Question

You want to use the cumulative distribution function

\[ P(X \leq k) = 1-(1-p)^k \]

to find the probability that an experiment requires \(30\) trials or less to succeed. What is the value of \(k\) ?

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Answer

\(30\).

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