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AI SL · 13.2

Probability Distributions

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13.2.1

Discrete random variables and distribution tables

13.2.1

Discrete random variables and distribution tables

Target: I can say what a discrete random variable is and write P(X = x) in words. (+4 more)

By the end of this section

Key points

Random variables and distributions

  • A discrete random variable X takes separate values, and which one it takes depends on chance. \(P(X=2)\) means the probability that X is 2.
  • The probability distribution lists every value of X with its probability, usually in a table.
  • The probabilities in a distribution add to 1: \(\sum P(X=x)=1\). Use this to find a missing probability k, and as a check on every table you build.
  • Every probability is between 0 and 1. If k comes out negative or bigger than 1, reject it.

Building a distribution

  • List every value X can take, then find \(P(X=x)\) for each. A tree diagram or a table of outcomes keeps the routes in order.
  • Several routes can give the same value of X. Add their probabilities.
  • Without replacement: after each draw the totals change, so the second probability is out of one fewer item.
  • A fair die with n faces is uniform: each score has probability \(\frac{1}{n}\).
  • Counting one outcome in 3 trials: every arrangement has the same probability, so multiply by the number of arrangements.

Cumulative and conditional probability

  • \(P(X\le a)\) means add the probabilities of every value up to and including a. \(P(X>a)=1-P(X\le a)\).
  • Conditional probability: \(P(A\mid B)=\frac{P(A\text{ and }B)}{P(B)}\). In a table, keep only the values where B is true. Their total is the new denominator.
  • Write \(P(Y=0\mid Y\le1)\) as P(Y = 0) over P(Y ≤ 1), not P(Y = 0) over 1.

Explore: a probability distribution table

Type the values X can take and the probability of each. Leave one probability empty and it is found for you.

x
P(X = x)
Total of the probabilities
Expected value, E(X)

Type a table to see its bars and its expected value.

  1. The marker under the axis is E(X). Change one probability at a time. Which way does the marker move, and why?
  2. Can E(X) be a value that X never takes? Try the fair die.
  3. In the game, change the prize so that the game is fair.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Pens taken without replacement

A box holds 6 blue pens and 4 green pens. Two pens are taken at random, one after the other, without replacement. X is the number of green pens taken. Find the probability distribution of X.

4/10G3/9GGG6/9BGB6/10B4/9GBG5/9BBB
Working
  1. No green pens

    \(P(X=0)=\frac{6}{10}\times\frac{5}{9}=\frac{30}{90}=\frac{1}{3}\)

    Blue, then blue. After one blue pen, 9 pens are left and 5 are blue.

  2. One green pen

    \(P(X=1)=\frac{4}{10}\times\frac{6}{9}+\frac{6}{10}\times\frac{4}{9}=\frac{48}{90}=\frac{8}{15}\)

    Two routes give one green pen: GB and BG. Add them.

  3. Two green pens

    \(P(X=2)=\frac{4}{10}\times\frac{3}{9}=\frac{12}{90}=\frac{2}{15}\)

    Green, then green.

  4. x012P(X = x)1/38/152/15
    The table

    \(\frac{1}{3}+\frac{8}{15}+\frac{2}{15}=1\)

    The probabilities add to 1, so the table is complete.

Video walkthrough coming
Finding k, then P(Y ≤ 1) and a conditional

The table shows the probability distribution of Y, the number of faults in a 100 m length of cable. (a) Find the value of k. (b) Find P(Y ≤ 1). (c) Find P(Y = 0 | Y ≤ 1).

y012more than 2P(Y = y)0.380.340.18k
Working
  1. Sum to 1

    \(0.38+0.34+0.18+k=1\Rightarrow k=0.1\)

    Every probability in the table adds to 1.

  2. Cumulative

    \(P(Y\le1)=P(Y=0)+P(Y=1)=0.38+0.34=0.72\)

    Y ≤ 1 means Y is 0 or 1.

  3. Conditional

    \(P(Y=0\mid Y\le1)=\frac{0.38}{0.72}=\frac{19}{36}=0.528\)

    Given that Y ≤ 1, only the first two columns count. Their total, 0.72, is the denominator.

Video walkthrough coming
Counting one number over three spins

A spinner has 5 equal sectors numbered 1 to 5. It is spun three times. X is the number of times it lands on 4. Find the probability distribution of X.

Working
  1. One spin

    \(P(4)=\frac15,\quad P(\text{not }4)=\frac45\)

    Each spin has the same two outcomes, and the spins do not affect each other.

  2. X = 0 and X = 3

    \(P(X=0)=\left(\frac45\right)^3=\frac{64}{125},\quad P(X=3)=\left(\frac15\right)^3=\frac{1}{125}\)

    Only one arrangement each.

  3. X = 1 and X = 2

    \(P(X=1)=3\times\frac15\times\left(\frac45\right)^2=\frac{48}{125},\quad P(X=2)=3\times\left(\frac15\right)^2\times\frac45=\frac{12}{125}\)

    A single 4 can come first, second or third: 3 arrangements, each with the same probability.

  4. x0123P(X = x)64/12548/12512/1251/125
    The table

    \(\frac{64}{125}+\frac{48}{125}+\frac{12}{125}+\frac{1}{125}=1\)

    Check that the total is 1.

Video walkthrough coming