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Discrete Random Variables and Probability Distributions

A random variable attaches a number to each outcome, and its distribution lists every value the variable can take with the probability of taking it. Those probabilities have to total 1.

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What a learner can do afterwards

  • Write the distribution of the total when two dice are added
  • Use the total of 1 to find a missing probability in a table
  • Find P(X ≥ 3) from a given distribution

1 · Read

The dice stall pays on the total of two fair dice rolled independently. Each pair has chance 1/6 × 1/6 = 1/36. A random variable is a rule assigning a number to each outcome. Here X adds the two faces. The rule is defined before rolling; the roll reveals its value. For equally likely outcomes, divide the count giving a value by the total count. For known unequal outcome probabilities, add the probabilities of outcomes giving that value.

Tap to fill the grid, one at a time: 6 rows of 6.
Thirty-six squares, one for each pair of faces, all equally likely. Every square carries a total, and several squares carry the same one.

Count the pairs giving each total. Only double one gives 2, so P(X = 2) = 1/36. Four pairs give 5: 1 and 4, 2 and 3, 3 and 2, 4 and 1, so P(X = 5) = 4/36. Listing each total from 2 to 12 with its chance gives the probability distribution of X. These probabilities describe the model. Observed relative frequencies are counts divided by the number of rolls. In a finite run they may be close to the model or far from it; exact agreement is possible.

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The number of pairs behind each total, climbing to six at a total of 7. Totals 10, 11 and 12 carry on down the other side at 3, 2 and 1.

Add those counts: 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36, so the probabilities add to 36 / 36 = 1. Every roll produces exactly one total, so the eleven totals cover every roll with no overlaps. Every distribution totals 1 and has no negative entries, so you can find a missing probability.

Try it together

A spinner stall has a random variable Y with values 1, 2, 3 and 4. Its probabilities are 0.2, 0.35, a smudge and 0.15. They total 1, so the smudge is 1 - 0.2 - 0.35 - 0.15 = 0.3. P(Y ≥ 3) means 3 or 4. One spin gives one score, so these events cannot both happen and their joint chance is 0. Add their probabilities: 0.3 + 0.15 = 0.45. At most 2 means adding the entries for 1 and 2 instead.

One missing entry from the total of 1, and P(Y at least 3) by adding upwards.

X is a rule, not a number. It assigns a number to each outcome. Its distribution lists every possible value with its probability, so it answers questions about the total. Unlisted values have probability zero; for two dice, 13 cannot occur. The table keeps values and probabilities, not the faces that produced a total of 6. Different games can share the same distribution.

A random variable turns an outcome into a number. Its distribution lists each value with its chance, built by counting. The chances total 1, which finds a missing one. For P(X ≥ k), add upwards.

Original lesson, question and flashcard revisions authored for this dataset.

The rewritten readings were written for this dataset. Includes original questions, references to Shafer and Zhang and OER Commons, and videos by Khan Academy. Third-party license terms are unknown.

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At the last stop, you learned that independent chances multiply. The dice stall runs on it: two fair dice, each roll independent, so 6 × 6 = 36 equally likely pairs of faces. The stall pays on the total of the two, not on which die showed what.

The stall calls the total X. What sort of thing is X before the dice are thrown?

Tap to fill the grid, one at a time: 6 rows of 6.

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Discrete Random Variables and Probability Distributions · Mathematics, ages 16 to 17 · LightMySky