Module 28 of 47Lesson 1 of 8 · Distributions 1 · Discrete random variables & probability distributions
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Probability distributions: discrete, binomial & normal · Essential understanding 1

Discrete random variables & probability distributions

IB AI SL 4.7with Prof. Kitso
💻 Using the TI-Nspire CX Student Software on a laptop
  • The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
  • menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
  • enter = Enter  ·  tab = Tab (next box)  ·  esc = Esc  ·  del = Backspace.
  • Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
  • Fraction: type /  ·  √: type sqrt(  ·  π: type pi  ·  eˣ: type exp(  ·  store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
  • New page: Ctrl+I (or the Insert menu)  ·  change page: Ctrl+← / Ctrl+→  ·  graph entry line: Ctrl+G  ·  var: Ctrl+L.
  • Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
  • IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
  • If you type nDeriv(, the software shows it as centralDiff( — it is the same command.
The big idea: A discrete random variable is a number that depends on chance and can only take separate values (0, 1, 2, …). Its probability distribution is a list of every possible value with its probability, and the probabilities always add up to 1.
By the end of this lesson you can:
  • explain what a discrete random variable is and write it as X
  • write a probability distribution as a table
  • use to find a missing probability or an unknown such as k
  • find probabilities such as P(X ≥ 2) and P(X < 3) by adding rows of the table
  • find a conditional probability from a table
  • build a distribution from a coin, a dice or a draw without replacement
🧑🏾 Imagine this…
Kabo tosses a fair coin 3 times and counts the heads. Before he starts, he does not know the answer. It could be 0, 1, 2 or 3, and it depends on chance. We give this unknown number a name, X. The question changes from “what will happen?” to “how likely is each possible value of X?” Writing those chances in a table is called finding the probability distribution of X.
Random variable
a quantity whose value depends on chance. We name it with a capital letter, such as X or Y.
Discrete random variable
a random variable that takes only separate values you can list or count, such as the number of heads, the number of goals or the score on a dice.
read “the probability that X is equal to x”. Capital X is the variable. Small x is one particular value, for example .
Probability distribution
all the possible values of X together with their probabilities, usually shown in a two-row table.
★ Two rules every probability distribution obeys
  • Every probability is between 0 and 1: .
  • All the probabilities add up to exactly 1:   (learn this one, you need it all the time).
  • To find P(X ≥ 2), add the probabilities of every value that is 2 or more. To find P(X < 3), add the values that are less than 3 (2, 1, 0, …).
  • Conditional probability (given that something has already happened): . Shrink the table to the values allowed by B, then divide by their total.
x0123
P(X = x)0.1250.3750.3750.125

That is the distribution for the number of heads in 3 tosses of a fair coin (Example 1.1 shows where the numbers come from). Check the rules: every entry is between 0 and 1, and 0.125 + 0.375 + 0.375 + 0.125 = 1.

⚠️ Watch out
  • The probabilities must add to exactly 1. If your table adds to 0.9 or 1.1, something is missing or wrong. If an unknown k comes out negative, or bigger than 1, re-check your adding.
  • Strict or not? P(X < 3) does not include 3, but P(X ≤ 3) does. When X can only be a whole number, P(X < 3) = P(X ≤ 2). Always ask yourself “is the end value included?”
  • Do not mix up x (the values in the top row) with P(X = x) (the probabilities in the bottom row). You add probabilities, never the x values, when you are finding a probability.
💡 Command terms
Write down the probability distribution = give the table of x and P(X = x). Show that a table is valid = add the probabilities and show the total is 1. Find k = form an equation from ΣP = 1 and solve it. In every case, give probabilities as fractions or to 3 significant figures.
🧪 Try it yourselfSee a distribution as bars
The bars show the distribution of the number of heads in 3 tosses (n = 3, p = 0.5). Tap the “from” and “to” boxes: the red bars are added together. Set from 0 to 3 and the total is 1. Set from 2 to 3 to see P(X ≥ 2). You will learn the name “binomial” in Lesson 3.

Worked examples

Example 1.1Building a distribution from outcomes
Kabo tosses a fair coin 3 times. Let X be the number of heads. (a) Write down the probability distribution of X. (b) Find P(X ≥ 2).
🧠 Think it through
  1. List all 8 equally likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
  2. Count the heads in each outcome. Three heads: 1 way. Two heads: 3 ways. One head: 3 ways. No heads: 1 way.
  3. Each probability is .
  4. Check that the four probabilities add to 1.
✍️ By hand
🔢 On your TI-Nspire CX II
  1. Check the total with sum( ) on a Calculator page. Type the four probabilities as a list in curly brackets, ctrl💻 Ctrl ( .
    1.1*DocRADsum({0.125,0.375,0.375,0.125})10.375+0.1250.5
✅ Answer (a) x = 0, 1, 2, 3 with P(X = x) = , , , . (b) P(X ≥ 2) = .
Writing the distribution as a table, and checking that the total is 1, is worth a mark on its own in many questions.
Example 1.2Finding a missing probability
The number of customers waiting at Neo's stall when she opens is a random variable X with this distribution:
x01234
P(X = x)0.1k0.30.250.05
  1. (a) Find the value of k.
  2. (b) Find P(X ≥ 2).
  3. (c) Find P(X < 3).
  4. (d) Find P(1 ≤ X ≤ 3).
🧠 Think it through
  1. The probabilities must add up to 1, so write the equation 0.1 + k + 0.3 + 0.25 + 0.05 = 1.
  2. (b) Add the rows for x = 2, 3, 4.
  3. (c) “Less than 3” means x = 0, 1, 2. The value 3 is not included.
  4. (d) Both ends are included: x = 1, 2, 3.
✍️ By hand
(b)
(c)
(d)
🔢 On your TI-Nspire CX II
  1. Let the calculator solve for k, then add the rows with sum( ). Use menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve for the equation.
    1.1*DocRADnSolve(0.1+k+0.3+0.25+0.05=1,k)0.3sum({0.3,0.25,0.05})0.6sum({0.1,0.3,0.3})0.7sum({0.3,0.3,0.25})0.85
✅ Answer (a) k = 0.3. (b) 0.6. (c) 0.7. (d) 0.85.
Check by a different route: P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − (0.1 + 0.3) = 0.6. Same answer.
Example 1.3An unknown in two or more rows
A random variable Y has this distribution:
y1234
P(Y = y)0.152k3k0.25
  1. (a) Find k, and write down P(Y = 2) and P(Y = 3).
  2. (b) Find the probability that Y = 2, given that Y ≥ 2.
🧠 Think it through
  1. Add everything and set the total equal to 1: the k terms collect together.
  2. Put k back in to get the probabilities of the two middle rows.
  3. “Given that Y ≥ 2” means only the rows y = 2, 3, 4 are allowed. Add them to get the bottom of the fraction.
  4. The top of the fraction is the row you want, y = 2.
✍️ By hand
(b)
🔢 On your TI-Nspire CX II
  1. Solve for k, then divide for the conditional probability.
    1.1*DocRADnSolve(0.15+2·k+3·k+0.25=1,k)0.120.240.850.282353
✅ Answer (a) k = 0.12, P(Y = 2) = 0.24, P(Y = 3) = 0.36. (b) 0.282 (3 s.f.).
In 2k, the 2 and k are multiplied. On the GDC type 2k or 2*k; both work.
Example 1.4A distribution given by a rule
A prize machine at a school fair gives a prize level X ∈ {1, 2, 3, 4, 5}, where . (a) Show that . (b) Write down the probability distribution. (c) Find P(X ≥ 4).
🧠 Think it through
  1. Put x = 1, 2, 3, 4, 5 into k(6 − x) to see the five probabilities in terms of k: 5k, 4k, 3k, 2k, k.
  2. They add to 1, so 15k = 1.
  3. Substitute k = into each row and simplify.
✍️ By hand
(c)
🔢 On your TI-Nspire CX II
  1. Use seq( to build the five terms, then solve for k.
    1.1*DocRADnSolve(sum(seq(k·(6−x),x,1,5))=1,k)0.0666667
✅ Answer (a) 15k = 1, so k = . (b) x = 1, 2, 3, 4, 5 with P = , , , , . (c) .
“Show that k = ” means you must write the equation 15k = 1 yourself. Stating k = from the GDC alone does not earn the marks.
Example 1.5A draw without replacement
A box holds 9 phones, of which 3 are faulty. Kabo takes 2 phones at random, one after the other, without putting any back. Let X be the number of faulty phones he takes.
  1. (a) Draw up the probability distribution of X.
  2. (b) Find P(X ≥ 1).
🧠 Think it through
  1. Without replacement, the second probability changes: after one phone is gone there are only 8 left.
  2. X = 0 means both are good: 6 good out of 9, then 5 good out of 8.
  3. X = 2 means both are faulty: 3 out of 9, then 2 out of 8.
  4. X = 1 is the rest. You can find it by subtraction, because the three probabilities must add to 1.
✍️ By hand
(b)
🔢 On your TI-Nspire CX II
  1. Type the products as decimals (press ctrl💻 Ctrl enter ↵💻 Enter to get a decimal).
    1.1*DocRAD69·580.41666739·280.0833331−0.416667−0.0833330.5
✅ Answer (a) x = 0, 1, 2 with P(X = x) = , , . (b) = 0.583 (3 s.f.).
Check by finding P(X = 1) directly: faulty then good, or good then faulty: 2 × ( × ) = . It matches.

Challenge questions

⭐ Challenge 1.6A table with two unknown-looking entriesPaper 1 style
The random variable X has this probability distribution:
x1234
P(X = x)0.15a0.252a
  1. (a) Find the value of a.
  2. (b) Find P(X > 2).
  3. (c) Find P(X = 4 | X > 2).
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Total = 1, so 0.15 + a + 0.25 + 2a = 1.
  2. “X > 2” means x = 3 and x = 4 only (2 is not included).
  3. Conditional: the row you want divided by the probability of the condition.
✍️ By hand
(b)
(c)
🔢 On your TI-Nspire CX II
  1. Check on a Calculator page.
    1.1*DocRADnSolve(0.15+a+0.25+2·a=1,a)0.20.40.650.615385
✅ Answer (a) a = 0.2. (b) 0.65. (c) = 0.615 (3 s.f.).
⭐ Challenge 1.7Larger score on two dicePaper 2 style
Two fair four-sided dice, each numbered 1 to 4, are rolled together. Let X be the larger of the two scores (if both scores are the same, X is that score).
  1. (a) List the 16 possible outcomes, or draw a table, and find the probability distribution of X.
  2. (b) Show that your probabilities add to 1.
  3. (c) Find P(X ≥ 3).
  4. (d) Find P(X = 4 | X ≥ 3).
  5. (e) Kabo says: “X = 4 is the most likely value, so it happens more than half the time.” Is he right? Use your table to answer.
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Draw a 4 × 4 grid of the 16 pairs (first dice, second dice). Fill each cell with the larger number.
  2. Count how many cells contain a 1, a 2, a 3 and a 4. Divide each count by 16.
  3. X ≥ 3 means X is 3 or 4.
  4. Conditional: P(X = 4) divided by P(X ≥ 3).
  5. For the last part, compare P(X = 4) with 0.5.
✍️ By hand
: only (1,1), so 1 outcome. : (1,2), (2,1), (2,2), so 3 outcomes.
: 5 outcomes. : 7 outcomes.
(a) for x = 1, 2, 3, 4.
(b)
(c)
(d)
(e)
🔢 On your TI-Nspire CX II
  1. Check the arithmetic.
    1.1*DocRADsum{1,3,5,7}16112160.757120.5833337160.4375
✅ Answer (a) x = 1, 2, 3, 4 with P = , , , . (c) . (d) = 0.583. (e) No: X = 4 is the most likely single value, but its probability is only = 0.4375, which is less than one half.
“Most likely value” (the mode of the distribution) does not mean “likely more often than not”. A value can have the biggest probability and still be less than 0.5.
Next: Lesson 2 · Expected value & fair games
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