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\[ \begin{align} \sigma &= \sqrt{ \frac{1-0.2}{0.2^2} } \\ &= \sqrt{20} \\ &= 4.472133. \end{align}\]

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\[ \begin{align} \sigma^2 &= \frac{1-\frac{1}{6}}{\left(\frac{1}{6}\right)^2} \\ &=\frac{\frac{5}{6}}{\frac{1}{36}} \\ &= 30. \end{align} \]

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.