Year 10 Standard · Unit 4 Sets, Venn diagrams and probability · Lesson 3 of 6

How do you list every outcome so that a probability is fair?

A probability is a share of a sample space, so the list of outcomes has to be complete and every outcome has to be equally likely. A table makes that list, and the complement rule saves counting twice.

Statistics and ProbabilityNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 8: write 3/12 in its simplest form.
    Answer1/4. Divide the top and the bottom by 3.
  2. From Year 7: write 7/20 as a percentage.
    Answer35%. 7/20 = 35/100.
  3. From Lesson 2: n(A) = 9, n(B) = 5 and n(A ∩ B) = 2. What is n(A ∪ B)?
    Answer12. 9 + 5 − 2 = 12.
Words used in this lesson
outcome
one possible result of an experiment. For example: the pair (2, 5) on two dice.
sample space
the list of all possible outcomes, written S. For example: the six faces 1 to 6 of one die.
event
a set of outcomes you are interested in. For example: the dice show a double.
equally likely
each outcome has the same chance of happening. For example: each side of a fair die.
probability
a number from 0 to 1 that says how likely an event is. For example: 1/6 for rolling a six.
complement
the event that A does not happen, written A′. For example: not rolling a six.

Eleven totals, so one in eleven?

At a market stall in Dakar a game costs a coin to play. You roll two fair dice and win if the numbers add up to 7. The stall owner says: “The total can be 2, 3, 4 and so on up to 12. That is eleven totals, so you have a one in eleven chance, and I pay ten times your coin if you win.”

Predict first: is the owner's number right? Are all eleven totals equally likely? If not, which total do you think can be made in the most ways, and why? Then use the first figure to count.

both even
9outcomes in the event
9/36 = 1/4P(event)

Both even: 9 of the 36 equally likely outcomes, so the probability is 9/36 = 1/4.

Show me the steps
  1. Both dice even: 3 choices on each die, so 9 of the 36 cells, and the probability is 9/36 = 1/4.
  2. Slide to sum is 7: 6 of the 36 cells, so the probability is 6/36 = 1/6. A total of 7 can be made more ways than any other total, and sum is 8 has only 5.
  3. Slide to doubles: the diagonal, 6 cells, so 6/36 = 1/6, the same as a sum of 7 although the cells are different.
  4. Slide to differ by 2: 8 cells in two parallel diagonals, so the probability is 8/36 = 2/9.

Probability is a share of a sample space whose outcomes are equally likely.

  • The sample space S is the list of every possible outcome. A table lists the outcomes of two stages without missing any.
  • An event A is the set of outcomes you are interested in. P(A) = n(A) ÷ n(S).
  • P(A) is always between 0 (impossible) and 1 (certain), and P(S) = 1.
  • The complement A′ is “A does not happen”: P(A′) = 1 − P(A).
9
0.375P(A)
0.625P(A′)

P(A) = 9/24 = 0.375 and P(A′) = 15/24 = 0.625. Together they make 24/24 = 1.

Show me the steps
  1. 9 tickets in A: P(A) = 9/24 = 0.375 and P(A′) = 15/24 = 0.625. Together they make 24/24 = 1.
  2. Slide to 0 tickets: A is impossible, P(A) = 0, and P(A′) = 24/24 = 1, so A′ is certain.
  3. Slide to 24 tickets: A is certain, P(A) = 1 and P(A′) = 0.
  4. Slide to 6 tickets: P(A) = 6/24 = 0.25 and P(A′) = 18/24 = 0.75. A probability can never be below 0 or above 1.

Count equally likely outcomes, not just outcomes. The eleven totals are not equally likely, so counting them tells you nothing. The thirty-six pairs of dice are equally likely, so count those. Two coins have four outcomes (HH, HT, TH, TT), not three, because HT and TH are different outcomes.

Worked example: an event and its complement

A fair spinner has 10 equal sectors numbered 1 to 10. Find P(prime) and P(not prime).

  1. The sample space has 10 equally likely outcomes.
  2. The primes are 2, 3, 5 and 7, so 4 outcomes are in the event: P(prime) = 4/10 = 2/5.
  3. Not prime is the complement: P(not prime) = 1 − 4/10 = 6/10 = 3/5.
  4. Check: 6 sectors (1, 4, 6, 8, 9, 10) are not prime, and 6/10 = 3/5.

Your go. The first step is given. Two fair spinners, each numbered 1, 2 and 3, are spun and the numbers are added. Find P(the sum is 4).

  1. Draw a 3 by 3 table of the first spinner against the second: 9 equally likely outcomes.
  2. Fill in the sum of each cell, then count the cells that show 4.
Check the answer

The cells with a sum of 4 are (1, 3), (2, 2) and (3, 1): 3 cells out of 9, so P = 3/9 = 1/3. A common slip is to say 1/5, because the sums 2 to 6 are five totals.

Check before you go on

1. Three fair coins are tossed. Someone says: “There are 4 outcomes (0, 1, 2 or 3 heads), so the probability of three heads is 1/4.” What is wrong?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Two fair four-sided dice, with faces 1 to 4, are rolled and the numbers are added. How many of the 16 equally likely outcomes give a total of 5?

3. A bag holds 9 red, 6 blue and 5 green counters, all the same size. One counter is taken at random. What is P(not red), as a decimal?

Level 2 · Building

4. Two ordinary six-sided dice are rolled. How many of the 36 outcomes have numbers that differ by at least 4?

5. A teacher picks one student at random from a class of 25 in Valparaiso. 9 students walk to school, 11 take the bus and the rest cycle. What is the probability, as a percentage, that the student does not cycle?

6. For an event A, P(A) = x + 0.2 and P(A′) = 3x. What is x?

Level 3 · Stretch

7. Two ordinary six-sided dice are rolled and the numbers are multiplied. What is the probability, as a decimal, that the product is even?

8. Two four-sided dice, with faces 1 to 4, are rolled. How many of the 16 outcomes have a larger number of exactly 3? (If both dice show 3, the larger number is 3.)

Something to think about

A spinner has 5 equal sectors, and a friend says: it is either red or not red, so P(red) must be 1/2. Explain what is wrong, and say what you would need to know to find P(red).

One way to think about it

Two outcomes are not necessarily equally likely. “Red” and “not red” are the two events, but they cover different numbers of sectors. P(red) = (number of red sectors) ÷ 5. With 1 red sector P(red) = 1/5, and with 3 it is 3/5; the probability is 1/2 only if the sample space were split evenly, which 5 sectors cannot do.

Exit ticket

9. A fair coin is tossed and an ordinary fair six-sided die is rolled. How many equally likely outcomes are there altogether?