International Baccalaureate · DP 2: Applications and Interpretation SL
Discrete Random Variables
I
Random Variables
I.1
Random Variables and Their Values
From Events to Numbers
An event is a set of outcomes and its probability a number. Many questions about a random experiment are about a number attached to each outcome rather than about an event: the gain in a game, the number of heads in three tosses, the sum of two dice. Such a number is not fixed in advance, it depends on the outcome that chance produces; it is what the next definition sets down under the name of random variable.
Definition — Random Variable
A random variable, denoted \(X\), is a function that assigns a numerical value to each outcome \(\omega\) of the sample space \(U\) (also written \(\Omega\)). We write this value as \(X(\omega)\). 
Example
Let \(X\) be the number of heads when tossing 2 fair coins: 
(red coin) and 
(blue coin). Find \(X(\textcolor{colordef}{H},\textcolor{colorprop}{T})\).
The outcome \((\textcolor{colordef}{H},\textcolor{colorprop}{T})\) means the red coin shows heads (H) and the blue coin shows tails (T). Since \(X\) counts heads, there’s 1 head. Thus, \(X(\textcolor{colordef}{H},\textcolor{colorprop}{T}) = 1\).
Definition — Discrete Random Variable
A random variable is discrete if its set of possible values is finite or countably infinite. This means we can list all possible values. Unless stated otherwise, the random variables considered take finitely many values; the rank of a first success, met later, is the exception. Definition — Events Involving a Random Variable
For a random variable \(X\): - \((X = x)\): The set of outcomes where \(X\) takes the value \(x\).
- \((X \leq x)\): The set of outcomes where \(X\) is less than or equal to \(x\).
- \((X \geq x)\): The set of outcomes where \(X\) is greater than or equal to \(x\).
Example
Let \(X\) be the number of heads when tossing 2 coins: and . List the outcomes for \((X = 0)\), \((X = 1)\), \((X = 2)\), \((X \leq 1)\), and \((X \geq 1)\). - \((X = 0) = \{(\textcolor{colordef}{T},\textcolor{colorprop}{T})\}\) (no heads).
- \((X = 1) = \{(\textcolor{colordef}{T},\textcolor{colorprop}{H}), (\textcolor{colordef}{H},\textcolor{colorprop}{T})\}\) (one head).
- \((X = 2) = \{(\textcolor{colordef}{H},\textcolor{colorprop}{H})\}\) (two heads).
- \((X \leq 1) = (X = 0) \cup (X = 1) = \{(\textcolor{colordef}{T},\textcolor{colorprop}{T}), (\textcolor{colordef}{T},\textcolor{colorprop}{H}), (\textcolor{colordef}{H},\textcolor{colorprop}{T})\}\) (at most one head).
- \((X \geq 1) = (X = 1) \cup (X = 2) = \{(\textcolor{colordef}{T},\textcolor{colorprop}{H}), (\textcolor{colordef}{H},\textcolor{colorprop}{T}), (\textcolor{colordef}{H},\textcolor{colorprop}{H})\}\) (at least one head).
Example
A quiz has three questions, each with four options and one correct answer. A student answers each question at random, independently of the others. Define a random variable that models the situation, give its possible values, and compute the probability that the student gets every answer wrong. - The variable: let \(X\) be the number of correct answers among the three. To each outcome, a sequence of three chosen options, \(X\) attaches the count of correct ones.
- Its possible values: \(0\), \(1\), \(2\) and \(3\).
- The event \((X = 0)\) is “the three answers are wrong”, and each answer is wrong with probability \(\frac{3}{4}\): $$\begin{aligned}[t] P(X = 0) &= \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} && \scriptscriptstyle\text{(independent answers)} \\ &= \frac{27}{64} \approx 0.42. \end{aligned}$$
Skills to practice
- Identifying the Possible Values
- Finding Events Involving a Random Variable
- Finding the Value of a Random Variable
I.2
Probability Distributions
Forgetting the Outcomes and Keeping the Values
Once the variable is defined, one often sets the outcomes aside and keeps only what \(X\) produces: its values, each with the probability of the event “\(X\) takes this value”. Recording these probabilities, as a table or as a formula, is what the next definition calls the probability distribution of \(X\). The outcomes come back whenever such a probability has to be computed.
Definition — Probability Distribution
The probability distribution of a random variable \(X\) lists the probability \(P(X = x_i)\) for each possible value \(x_1,x_2,\dots,x_n\). Here each \(x_i\) is a possible value of \(X\) — a real number — not an outcome. It can be shown as a table or formula. Proposition — Characteristic of a Probability Distribution
For a random variable \(X\) with possible values \(x_1,x_2,\dots,x_n\), we have - \(0 \leq P(X=x_i) \leq 1\) for all \(i=1,\dots,n\),
- \(\displaystyle\sum_{i=1}^n P(X=x_i) =P(X=x_1)+P(X=x_2)+\dots+P(X=x_n)= 1 \).
Example
Let \(X\) be the number of heads when tossing, independently, 2 fair coins: and . - List the possible values of \(X\).
- Find the probability distribution.
- Create the probability table.
- Draw the probability distribution graph.
- Possible values: \(0\) (no heads), \(1\) (one head), \(2\) (two heads).
- Probability distribution:
- \(P(X = 0) = P(\{(\textcolor{colordef}{T},\textcolor{colorprop}{T})\}) = \frac{1}{4}\),
- \(P(X = 1) = P(\{(\textcolor{colordef}{T},\textcolor{colorprop}{H}), (\textcolor{colordef}{H},\textcolor{colorprop}{T})\}) = \frac{2}{4} = \frac{1}{2}\),
- \(P(X = 2) = P(\{(\textcolor{colordef}{H},\textcolor{colorprop}{H})\}) = \frac{1}{4}\).
- Probability table:
\(x\) 0 1 2 \(P(X = x)\) \(\frac{1}{4}\) \(\frac{1}{2}\) \(\frac{1}{4}\) - Graph:
Skills to practice
- Checking that a Distribution Is Valid
- Finding a Probability Distribution
- Constructing a Probability Distribution
II
Expected Value
II.1
Expected Value
The expected value of a random variable \(X\) is the "average you’d expect if you repeated the experiment many times". It’s found by taking all possible values, multiplying each by its probability, and adding them up — essentially a weighted average where the probabilities act as the weights.
Definition — Expected Value
For a random variable \(X\) with possible values \(x_1, x_2, \ldots, x_n\), the expected value, \(E(X)\), also called the mean, is:$$\begin{aligned}E(X) &= \sum_{i=1}^{n} x_i P(X = x_i)\\
&= x_1 P(X = x_1) + x_2 P(X = x_2) + \cdots + x_n P(X = x_n)\\
\end{aligned}$$ Example
You toss 2 fair coins, and \(X\) is the number of heads. The probability distribution is: | \(x\) | 0 | 1 | 2 |
| \(P(X = x)\) | \(\frac{1}{4}\) | \(\frac{1}{2}\) | \(\frac{1}{4}\) |
Calculate \(E(X)\) using the formula:$$\begin{aligned}E(X) &= 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 2 \times \frac{1}{4} \\
&= \frac{1}{2} + \frac{2}{4} \\
&= 1\end{aligned}$$ So, on average, you expect 1 head when tossing 2 coins.
Fair Game
When \(X\) is the net gain of a player, stake deducted, \(E(X)\) is the average gain per play over a large number of plays. A game is fair when \(E(X) = 0\): on average the player neither wins nor loses. If \(E(X) > 0\), the game favours the player; if \(E(X) < 0\), it favours the organiser. With \(G\) the gross gain and \(m\) the stake, \(X = G - m\): \(G - m\) takes each value of \(G\) decreased by \(m\), so its weighted mean decreases by \(m\), and \(E(X) = E(G) - m\). The game is fair when the stake equals the expected value of the gross gain.
Example
A fair die is rolled. The player receives \(6\) euros for a \(6\), \(3\) euros for a \(5\) and nothing otherwise. Determine the stake \(m\) that makes the game fair.
The gross gain \(G\) takes the values \(6\), \(3\) and \(0\) with probabilities \(\frac{1}{6}\), \(\frac{1}{6}\) and \(\frac{4}{6}\). $$\begin{aligned}[t] E(G) &= 6 \times \frac{1}{6} + 3 \times \frac{1}{6} + 0 \times \frac{4}{6} = 1.5 && \scriptscriptstyle\text{(definition of the expected value)} \\
E(X) &= E(G - m) = E(G) - m = 1.5 - m && \scriptscriptstyle\text{(each value of } G \text{ decreased by } m\text{)} \end{aligned}$$ The game is fair when \(E(X) = 0\), that is \(m = 1.5\): a stake of \(1.50\) euros. With a stake of \(2\) euros, \(E(X) = -0.5\): the player loses \(0.50\) euro per play on average.
Skills to practice
- Calculating Expected Values
- Finding the Stake of a Fair Game
- Interpreting Expected Values
III
Repeated Trials
III.1
Sequences of Independent Trials
The Same Urn, Two Draws
An urn holds three red balls and two blue balls, indistinguishable by touch. A ball is drawn, its colour is noted, then a second ball is drawn. 
An urn: three red balls, two blue balls
An urn: three red balls, two blue balls
- With replacement, the first ball goes back into the urn: the second draw has the same probabilities as the first, \(\frac{3}{5}\) for red and \(\frac{2}{5}\) for blue, whatever the first result. The two draws do not influence each other: they are independent. The probability of the outcome (red, red) is \(\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\).
- Without replacement, the first ball stays out: the probabilities of the second draw depend on the first result, \(\frac{2}{4}\) for red after a red ball, \(\frac{3}{4}\) after a blue one. The two draws are not independent; the branches carry conditional probabilities, and the probability of (red, red) is \(\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}\).
Definition — Sequence of Trials
A sequence of \(n\) trials is a random experiment made of \(n\) random experiments performed one after the other, with sets of outcomes \(E_1, E_2, \dots, E_n\). An outcome of the sequence is the list \((x_1, x_2, \dots, x_n)\) of the results, with \(x_i \in E_i\); the sample space is the Cartesian product \(E_1 \times E_2 \times \dots \times E_n\) of all these lists. The sequence is represented by a tree with \(n\) levels, one per trial: a path from the root to a leaf is an outcome. Definition — Independent Trials: the Product Model
The \(n\) trials are independent when the result of any of them does not depend on the results of the others: on the tree, each outcome of a trial keeps the same probability whatever the results of the previous trials, so, at a given level of the tree, the branches leading to the same outcome carry the same weight whatever the node they leave from (the weights of the different outcomes of one trial need not be equal). In this model, the probability of an outcome is the product of the probabilities of its components: $$P\bigl((x_1, x_2, \dots, x_n)\bigr) = P_1(x_1) \times P_2(x_2) \times \dots \times P_n(x_n),$$ where \(P_i(x_i)\) is the probability of the result \(x_i\) in the \(i\)-th trial. For two trials, this is the multiplication rule for independent events.
A Model, Not a Theorem
Independence is an assumption about the situation, chosen when the trials do not influence each other: draws with replacement, repeated tosses of a coin, repeated shots of a player whose form does not change. In a given situation it is a modelling choice, stated rather than deduced from the data: the product formula then defines the probability model of the sequence. The independence is mutual: it concerns all the trials together, not only two by two.
Method — Computing with a Tree
- Draw the tree, one level per trial, and write the probability of each branch; the probabilities of the branches leaving a node add up to \(1\).
- Probability of a path: multiply the probabilities along the path, from the root to the leaf.
- Probability of an event: add the probabilities of the paths that realise it (two different paths are disjoint outcomes).
- "At least one": go through the complement "none", which has far fewer paths (a single one when each trial is reduced to "the outcome occurs or not").
Example
A fair six-sided die is rolled three times. The rolls are independent. Compute the probability of getting three sixes, exactly one six, and at least one six.
Each roll has two outcomes of interest, "\(6\)" with probability \(\frac{1}{6}\) and "\(\overline{6}\)" with probability \(\frac{5}{6}\); the tree has \(2^3 = 8\) paths. 
Three sixes: one path, \(\left(\frac{1}{6}\right)^3 = \frac{1}{216}\). Exactly one six: three paths (\(6\overline{6}\overline{6}\), \(\overline{6}6\overline{6}\), \(\overline{6}\overline{6}6\)), each of probability \(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{25}{216}\), so \(3 \times \frac{25}{216} = \frac{75}{216} = \frac{25}{72}\). At least one six: the complement of "no six", the path \(\overline{6}\overline{6}\overline{6}\) of probability \(\left(\frac{5}{6}\right)^3 = \frac{125}{216}\), so \(1 - \frac{125}{216} = \frac{91}{216}\).
Skills to practice
- Modelling a Sequence of Trials with a Tree
III.2
Repeating a Bernoulli Trial
Definition — Bernoulli Trial
A Bernoulli trial is a random experiment with exactly two outcomes, called success (\(S\)) and failure (\(F\)). The probability of success is a number \(p\) strictly between \(0\) and \(1\) — a success that is neither impossible nor certain; the probability of failure is then \(1 - p\). 
Repeating the trial. The same Bernoulli trial is now repeated \(n\) times, the repetitions being identical (the probability of success is \(p\) at every trial) and independent (whatever the outcomes of the previous trials, the probability of success at the next one is still \(p\)) — this repetition is also called a Bernoulli scheme. An outcome of the whole experiment is a list of \(n\) letters \(S\) or \(F\), one per trial, and the tree of the experiment has \(2^n\) paths: \(4\) for two trials, \(8\) for three, \(16\) for four. By independence, the probability of a path is the product of the probabilities written on its branches: a path with \(k\) successes and \(n - k\) failures, in any order, has probability \(p^k (1 - p)^{n - k}\).
Method — Computing a Probability with the Tree
For \(n \leq 4\) repetitions of a Bernoulli trial: - Draw the tree, one level per trial, with \(p\) on every success branch and \(1 - p\) on every failure branch (the same numbers at every level, since the trials are identical and independent).
- Select the paths that make up the event: for instance “exactly two successes” selects the paths that contain exactly two letters \(S\).
- Multiply along each path, then add the selected paths: all the paths with the same number of successes have the same probability, so it is enough to count them.
Example
A basketball player scores a free throw with probability \(0.7\). She shoots three free throws, independently. - Draw the tree of the experiment.
- Compute the probability that she scores exactly two of the three.
- Compute the probability that she scores at least one.
- Each throw is a Bernoulli trial with success \(S\) (“scored”) of probability \(0.7\) and failure \(F\) of probability \(0.3\); three identical independent trials give a tree with \(8\) paths.
- “Exactly two scored” is made of the three paths \(SSF\), \(SFS\) and \(FSS\), each of probability \(0.7 \times 0.7 \times 0.3 = 0.147\): \(P = 3 \times 0.147 = 0.441\).
- “At least one scored” is the complement of “none scored”, the single path \(FFF\) of probability \(0.3^3 = 0.027\): \(P = 1 - 0.027 = 0.973\).
Example
A fair coin is tossed four times, the tosses being independent. Without drawing the whole tree, compute the probability of getting exactly two heads.
Each toss is a Bernoulli trial with success “heads” of probability \(\dfrac{1}{2}\); the four tosses are identical and independent, and the tree has \(2^4 = 16\) paths, each of probability \(\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}\). The paths with exactly two heads are those where the two letters \(S\) occupy two of the four positions: \(SSFF\), \(SFSF\), \(SFFS\), \(FSSF\), \(FSFS\), \(FFSS\) — six paths. Hence \(P = \dfrac{6}{16} = \dfrac{3}{8} = 0.375\). Listing the paths replaces the drawing: with four trials the tree is already wide, and beyond four the count of paths needs a tool of its own.
Skills to practice
- Recognising a Bernoulli Trial
- Computing with the Tree
- Solving Problems with the Tree
IV
The Bernoulli and Binomial Distributions
IV.1
The Bernoulli Distribution
A Bernoulli trial produces a random variable: \(X = 1\) if the trial is a success, \(X = 0\) if it is a failure. Its distribution is the Bernoulli distribution with parameter \(p\), the probability of success.
Definition — Bernoulli Distribution
A random variable \(X\) follows a Bernoulli distribution with parameter \(p\) if: - Possible values are 0 and 1.
- \(P(X = 1) = p\) and \(P(X = 0) = 1 - p\).
Example
A basketball player has an 80\(\pourcent\) chance of making a free throw. Let \(X = 1\) if the shot is made, and \(X = 0\) if it’s missed. - Does \(X\) follow a Bernoulli distribution?
- Find the probability of success.
- Yes, \(X\) has values 0 or 1, so it follows a Bernoulli distribution.
- Probability of success: \(P(X = 1) = 80\pourcent = 0.8\).
Example
A machine fills bottles. A bottle is rejected with probability \(0.05\). Let \(X = 1\) if the bottle is rejected and \(X = 0\) otherwise. - Give the probability distribution of \(X\).
- Compute \(E(X)\) and interpret it.
- More generally, \(X\) follows a Bernoulli distribution with parameter \(p\). Compute \(E(X)\).
- \(X\) follows a Bernoulli distribution with parameter \(p = 0.05\):
\(x\) \(0\) \(1\) \(P(X = x)\) \(0.95\) \(0.05\) - \(\begin{aligned}[t]E(X) &= 0 \times 0.95 + 1 \times 0.05 \\ &= 0.05\end{aligned}\)
Over a large number of bottles filled independently, the mean of the values of \(X\) — that is, the frequency of the \(1\), the proportion of rejected bottles — is expected to be close to \(0.05\). - \(\begin{aligned}[t]E(X) &= 0 \times (1 - p) + 1 \times p \\ &= p\end{aligned}\)
The expectation of a Bernoulli variable is its parameter: \(E(X) = p\).
The variance measures how spread out the values of a random variable are from its expected value. The standard deviation is the square root of the variance, giving a sense of typical deviation in the same units as \(X\).
Definition — Variance and Standard Deviation
The variance, denoted \(V(X)\), is:$$\begin{aligned}V(X) &= \sum_{i=1}^{n} (x_i - E(X))^2 P(X = x_i)\\
&= \left(x_1-E(X)\right)^2 P(X = x_1) + \left(x_2-E(X)\right)^2 P(X = x_2) + \cdots + \left(x_n-E(X)\right)^2 P(X = x_n)\\
\end{aligned}$$ The standard deviation, denoted \(\sigma(X)\), is \(\sigma(X) = \sqrt{V(X)}\). Example
You toss 2 fair coins, and \(X\) is the number of heads. The probability table is: | \(x\) | 0 | 1 | 2 |
| \(P(X = x)\) | \(\frac{1}{4}\) | \(\frac{1}{2}\) | \(\frac{1}{4}\) |
Calculate \(V(X)\):$$\begin{aligned}V(X) &= (0 - 1)^2 \times \frac{1}{4} + (1 - 1)^2 \times \frac{1}{2} + (2 - 1)^2 \times \frac{1}{4} \\
&= 1 \times \frac{1}{4} + 0 \times \frac{1}{2} + 1 \times \frac{1}{4} \\
&= \frac{1}{4} + 0 + \frac{1}{4} \\
&= \frac{1}{2} \\
\end{aligned}$$ The variance is \(\frac{1}{2}\).
Proposition — Expectation and Variance of a Bernoulli Distribution
For a Bernoulli random variable \(X\) with a probability of success \(p\), the following hold: - The expected value is \(E(X) = p\),
- The variance is \(V(X) = p(1 - p)\),
- The standard deviation is \(\sigma(X) = \sqrt{p(1 - p)}\).
- \(\begin{aligned}[t]E(X)&=0\times P(X=0)+1\times P(X=1)\\&=0\times(1-p)+1\times p\\&=p\end{aligned}\)
- \(\begin{aligned}[t] V(X) &= (0-p)^2(1-p) + (1-p)^2 p \\ &= p^2(1-p) + p(1-p)^2 \\ &= p(1-p) [p + (1-p)] \\ &= p(1-p) \\ \end{aligned} \)
Skills to practice
- Identifying Bernoulli Distributions
- Defining Bernoulli Trials and Variables
IV.2
The Binomial Distribution
Definition — Bernoulli Scheme
A Bernoulli scheme with parameters \(n\) and \(p\) is the sequence of \(n\) Bernoulli trials that are identical (same probability of success \(p\)) and independent: the same trial repeated \(n\) times, without any trial influencing the others. It is recognised by four conditions: - a fixed number \(n\) of trials;
- each trial has two outcomes, success and failure;
- the probability of success \(p\) is the same at every trial;
- the trials are independent.
Definition — Binomial Distribution
In a Bernoulli scheme with parameters \(n\) and \(p\), let \(X\) be the number of successes. \(X\) takes the values \(0, 1, \dots, n\); its probability distribution is called the binomial distribution with parameters \(n\) and \(p\), written \(\mathcal{B}(n, p)\), and we write \(X \sim \mathcal{B}(n, p)\). Proposition — Probability of \(k\) Successes
Let \(X \sim \mathcal{B}(n, p)\). The probability distribution of \(X\) is: 
Example
A basketball player has an 80\(\pourcent\) chance of making a free throw and takes 5 shots; the shots are assumed to be independent of one another. Let \(X\) be the number of shots made. - Does \(X\) follow a binomial distribution?
- Find the probability of making 4 shots.
- Yes, \(X\) follows a binomial distribution because it counts the number of successes (shots made) in 5 independent trials (free throws), each with a constant success probability of 0.8.
- As \(X \sim \mathcal{B}(5, 0.8)\), $$ \begin{aligned} P(X = 4) &= \binom{5}{4} (0.8)^4 (1-0.8)^1 \\ &= 5 \times 0.4096 \times 0.2 \\ &= 0.4096 \end{aligned} $$ The probability of making 4 out of 5 shots is \(0.4096 = 40.96\pourcent\).
Proposition — Expectation and Variance of a Binomial Distribution
For \(X \sim \mathcal{B}(n, p)\): - \(E(X) = n p\) (expected value),
- \(V(X) = n p (1 - p)\) (variance),
- \(\sigma(X) = \sqrt{n p (1 - p)}\) (standard deviation).
Example
A basketball player has an 80\(\pourcent\) chance of making a free throw and takes 5 shots; the shots are assumed to be independent of one another. Find the expected value and the standard deviation of the number of successful shots.
Let \(X\) be the number of successful shots. Since each shot is independent and has a success probability of 0.8, we have \(X \sim \mathcal{B}(5, 0.8)\).$$\begin{aligned}E(X) &= 5 \times 0.8 = 4, \\
V(X) &= 5 \times 0.8 \times (1 - 0.8) = 5 \times 0.8 \times 0.2 = 0.8, \\
\sigma(X) &= \sqrt{0.8} \approx 0.89.\end{aligned}$$ The expected value is 4 successful shots, the standard deviation about 0.89.
Example
- A machine makes defective parts with probability \(0.02\), independently from one part to the next. In a batch of \(150\) parts, \(X\) counts the defective ones. Give the distribution of \(X\), then \(E(X)\), and interpret it.
- A seed germinates with probability \(0.85\). Forty seeds are sown, independently. How many germinated seeds can be expected on average?
- Each part is a Bernoulli trial (success: “defective”, \(p = 0.02\)), repeated \(150\) times independently: \(X \sim \mathcal{B}(150, 0.02)\). Then \(E(X) = 150 \times 0.02 = 3\): over many batches, a batch contains on average \(3\) defective parts.
- \(X \sim \mathcal{B}(40, 0.85)\) and \(E(X) = 40 \times 0.85 = 34\): about \(34\) germinated seeds on average.
The Bar Chart of a Binomial Distribution
The bar chart of \(X \sim \mathcal{B}(n, p)\) shows the probabilities \(P(X = k)\) for \(k = 0, 1, \dots, n\). Its highest bar stands close to the expected value \(E(X) = np\). When \(np\) and \(n(1-p)\) are not too small, the chart is roughly bell-shaped and centred on \(np\): the bars rise, peak near \(np\), then fall roughly symmetrically. When \(np\) (or \(n(1-p)\)) is small, which happens for a given \(n\) when \(p\) is close to \(0\) (or to \(1\)), the chart loses its symmetry: it is pushed against the edge, its tallest bars stand at the very start (or the very end) and it falls quickly on one side only. Below, \(\mathcal{B}(20, 0.6)\) with \(E(X) = 12\), and \(\mathcal{B}(20, 0.05)\) with \(E(X) = 1\). 
Reading a bar chart the other way round gives information about the parameters: the abscissa of the highest bar is close to \(np\), so it gives an estimate of \(n\) when \(p\) is known, or of \(p\) when \(n\) is known.
Skills to practice
- Identifying Binomial Distributions
- Computing Binomial Probabilities and Expected Values
- Computing the Variance of a Binomial Distribution
- Identifying and Applying the Binomial Distribution
IV.3
Binomial Probabilities with Technology
Method — Cumulative Binomial Distribution
To calculate binomial probabilities, we use a graphing calculator: its probability function (PDF) gives \(P(X = k)\), and its Cumulative Distribution Function (CDF) gives probabilities of the form \(P(X \leq k)\). - On TI: Use
binompdf(n, p, k)for \(P(X = k)\) andbinomcdf(n, p, k)for \(P(X \leq k)\). - On Casio: Use
BinomialPD(k, n, p)andBinomialCD(k, n, p). - On NumWorks: Use the
Probabilityapplication, selectBinomial.
- \(P(X < k) = P(X \leq k-1)\)
- \(P(X \geq k) = 1 - P(X \leq k-1)\)
- \(P(X > k) = 1 - P(X \leq k)\)
- \(P(a \leq X \leq b) = P(X \leq b) - P(X \leq a-1)\)
Example
Consider a random variable \(X\) that follows the binomial distribution with parameters \(n = 100\) and \(p = 0.78\). Calculate the following probabilities. Round your results to three decimal places. $$P(X < 75), P(X > 79), P(X \geq 74), P(73 < X \leq 81)$$
With \(X \sim \mathcal{B}(100, 0.78)\), we use the calculator's cumulative distribution function:
- \(P(X<75) = P(X\leq 74) \approx \mathbf{0.197}\).
- \(P(X>79) = 1 - P(X\leq 79) \approx \mathbf{0.366}\).
- \(P(X\geq 74) = 1 - P(X\leq 73) \approx \mathbf{0.861}\).
- \(P(73
Method — Finding a Threshold
To find the smallest integer \(k\) such that \(P(X \leq k) \geq 1-\alpha\) (where \(\alpha\) is a small value like \(0.05\)), we use the Inverse Binomial function or a table of values on the calculator. Example
Let \(X \sim \mathcal{B}(50, 0.4)\). Find the smallest integer \(k\) such that \(P(X \leq k) \geq 0.95\).
Using the calculator's table of values:
- \(P(X \leq 24) \approx 0.9022\)
- \(P(X \leq 25) \approx 0.9427\)
- \(P(X \leq 26) \approx 0.9686\)
Skills to practice
- Computing Cumulative Binomial Probabilities with Technology
- Finding a Threshold
V
Going further
Skills to practice
- Solving Problems
- Quiz
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