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AI HL · 13.1

Probability

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13.1.8

Mixed practice: choosing the right tool

13.1.8

Mixed practice: choosing the right tool

Target: I can choose between a tree diagram, a Venn diagram, a two-way table and a formula for a probability question. (+5 more)

By the end of this section

Key points

Choosing the tool

  • Overlapping groups with counts or totals: draw a Venn diagram. Fill the overlap first.
  • A sequence of choices or events: draw a tree diagram. Multiply along a branch, add the branches that give the outcome.
  • Two categories counted together: a two-way table. Read totals from the margins.
  • Probabilities given as P(A), P(B), P(A ∩ B), P(A ∪ B): use the formulas.
  • Reading a tree backwards: P(first | second) = P(first and second) ÷ P(second). The bottom is the total of every path that ends in the second event.

Rules to remember

  • Without replacement: after each draw, both the number of items and the number of one colour can change.
  • Total probability: add the probabilities of every branch that ends in the outcome you want.
  • Conditional probability: \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). From a table, divide by the total of the row or column you are given.
  • Test for independence: compare \(P(A\cap B)\) with \(P(A)\times P(B)\), or compare \(P(A|B)\) with \(P(A)\).

Formulas for combined events

  • \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
  • \(P(A\cap B)=P(A)\times P(B|A)\).
  • \(P(B')=1-P(B)\): a complement given in the question is a value of \(P(B)\) in disguise.
  • For the sum of three dice, list the sets of three scores that give the sum, then count the orders of each set.

Explore: a two-stage tree diagram

Fill the tree from a bag of counters, with or without replacement, or type your own probabilities in the orange boxes. The grey box on each pair is 1 minus its partner.

The tree comes from

The four end products

Read out

Total of the four end products

Choose a bag or type your own probabilities.

  1. Choose "without replacement". Compare the second-stage branches after a red and after a blue. Why are they different? Then choose "with replacement": what changes?
  2. Look at the four end products. Why must they add to 1?
  3. Which is bigger, P(exactly one) or P(at least one)? Can they ever be equal? Try the counts to see.
  4. For "first, given second", why is the bottom of the fraction the probability of the second counter and not the first branch?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Two draws, different colours

A bag holds 7 red and 5 blue counters. Two counters are taken out at random, one after the other, without replacement. Find the probability that they are different colours.

Working
  1. Two ways

    Red then blue, or blue then red

    Different colours can happen in two orders.

  2. Red then blue

    \(\frac{7}{12}\times\frac{5}{11}=\frac{35}{132}\)

    Without replacement: 11 counters left, 5 of them blue.

  3. Blue then red

    \(\frac{5}{12}\times\frac{7}{11}=\frac{35}{132}\)

    The same by symmetry.

  4. Add

    \(P=\frac{70}{132}=\frac{35}{66}\)

    Add the two branches.

Video walkthrough coming
A two-way table and independence

The table shows how many students in Year 10 and Year 11 play football for the school. (a) Find the probability that a student plays football, given that they are in Year 10. (b) Find the probability that a student plays football, given that they are in Year 11. (c) Determine whether playing football is independent of year group.

PlaysDoes not playYear 102030Year 112416
Working
  1. Year 10 row

    Row total: 20 + 30 = 50. \(P=\frac{20}{50}=0.4\)

    The condition picks the row.

  2. Year 11 row

    Row total: 24 + 16 = 40. \(P=\frac{24}{40}=0.6\)

    Divide by this row's total.

  3. Compare

    \(0.4\ne 0.6\)

    If playing football were independent of year, the two would be equal.

  4. Conclude

    Not independent

    Year 11 students are more likely to play football.

Video walkthrough coming
A Venn diagram with a missing count

Of 30 families, 17 own a dog, 6 own a dog and a cat, and 4 own neither. (a) Find the number of families that own a cat but not a dog. (b) Find the probability that a family owns a cat. (c) Given that a family owns a cat, find the probability that they also own a dog.

11694DogCat
Working
  1. Fill in

    Dog only: 17 − 6 = 11. Neither: 4.

    The overlap and the outside are known.

  2. Cat only

    30 − 11 − 6 − 4 = 9

    Total minus the other three regions. Part (a): 9.

  3. Owns a cat

    \(P=\frac{15}{30}=\frac{1}{2}\)

    Cat only plus both: 9 + 6 = 15.

  4. Given a cat

    \(P=\frac{6}{15}=\frac{2}{5}\)

    Only the 15 cat owners count.

Video walkthrough coming
A tree with three equal branches

Dara goes to college by train, by bus or by bike, each with probability \(\frac{1}{3}\). The probability that Dara is late is 0.1 by train, 0.25 by bus and 0.05 by bike. Find the probability that Dara is late.

1/3Train0.1Late0.9On time1/3Bus0.25Late0.75On time1/3Bike0.05Late0.95On time
Working
  1. Multiply along branches

    Train, late: \(\frac{1}{3}\times 0.1\). Bus, late: \(\frac{1}{3}\times 0.25\). Bike, late: \(\frac{1}{3}\times 0.05\)

    Each branch ending in 'late' is a way to be late.

  2. Add the branches

    \(P=\frac{1}{3}(0.1+0.25+0.05)=\frac{1}{3}\times 0.4\)

    Take out the common factor \(\frac{1}{3}\).

  3. Result

    \(P=\frac{2}{15}\)

    Exact fraction; about 0.133.

Video walkthrough coming
The sum of three dice

Three fair six-sided dice are thrown. Find the probability that the sum of the scores is 9.

Working
  1. List the sets

    1 + 2 + 6; 1 + 3 + 5; 1 + 4 + 4; 2 + 2 + 5; 2 + 3 + 4; 3 + 3 + 3

    Write each set of three scores from smallest to largest.

  2. Count the orders

    1 + 2 + 6: 6; 1 + 3 + 5: 6; 1 + 4 + 4: 3; 2 + 2 + 5: 3; 2 + 3 + 4: 6; 3 + 3 + 3: 1; total 25

    Three different scores: 6 orders. Two the same: 3. All the same: 1.

  3. Probability

    \(P=\frac{25}{6^3}=\frac{25}{216}\)

    There are 216 equally likely outcomes. About 0.116.

Video walkthrough coming
Finding P(A ∩ B) from a formula

Events A and B satisfy \(P(A)=0.45\), \(P(A\cup B)=0.8\) and \(P(B')=0.6\). Find \(P(A\cap B)\).

Working
  1. The complement

    \(P(B)=1-0.6=0.4\)

    P(B′) is given. Turn it into P(B) first.

  2. The formula

    \(0.8=0.45+0.4-P(A\cap B)\)

    Use \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

  3. Rearrange

    \(P(A\cap B)=0.45+0.4-0.8=0.05\)

    Check: this must not be larger than either P(A) or P(B).

Video walkthrough coming
A tree read backwards

Machine P makes 70% of the parts in a factory and machine Q makes the rest. 2% of the parts from P are faulty and 5% of the parts from Q are faulty. A part is chosen at random and is found to be faulty. Find the probability that it was made by machine Q. Give your answer to 3 significant figures.

0.7P0.02Faulty0.0140.98OK0.6860.3Q0.05Faulty0.0150.95OK0.285
Working
  1. Multiply along the branches

    \(P(\text{P, faulty})=0.7\times0.02=0.014\) \(P(\text{Q, faulty})=0.3\times0.05=0.015\)

    The condition "faulty" is on the second stage, so use both branches that end in faulty.

  2. P(faulty)

    \(0.014+0.015=0.029\)

    This is the bottom of the fraction: every way to be faulty.

  3. Conditional probability

    \(P(Q\mid\text{faulty})=\dfrac{P(Q\cap\text{faulty})}{P(\text{faulty})}=\dfrac{0.015}{0.029}=0.517\)

    Use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\) with the whole tree behind it.

Video walkthrough coming