Math Vale

🎲 Probability Marsh · Probability

Compound Events

Work out probabilities for experiments with more than one stage, and keep track of whether the first stage changes the second.

In short

  • Stages of a compound event multiply, whether or not they are independent — what changes is the numbers, not the operation.
  • Without replacement, both the favourable count and the total drop for the next draw, and the answer is always smaller than with replacement.
  • A grid or a tree makes the sample space explicit, and its cells are equally likely even when the totals it produces are not.
  • P(at least one) = 1 - P(none), which turns a long list of cases into a single multiplication.

Two stages, one probability

A compound event takes more than one step: two draws from a bag, three tosses of a coin, a spinner and then a die. The whole skill rests on one question asked at every step:

Does what already happened change what happens next?

If it does not, the stages are independent and each keeps its own probability. If it does, they are dependent, and the later probabilities have to be adjusted.

Either way the stages multiply. What changes between the two cases is not the operation, only the numbers you feed into it.

With replacement and without

A jar holds 4 gold and 6 plain tokens, 10 in all.

With replacement. Draw one, note it, put it back. The jar is exactly as it was, so the second draw faces the same 4 out of 10:

P(both gold) = 4/10 x 4/10 = 16/100 = 4/25.

Without replacement. Draw one and keep it. If the first was gold, only 3 gold remain, out of only 9 tokens:

P(both gold) = 4/10 x 3/9 = 12/90 = 2/15.

Notice that both numbers change. Dropping the numerator but leaving the denominator at 10 is a half-done adjustment; the token has left the jar entirely, so it has to leave the total as well. And without replacement always gives the smaller answer, which is a quick check worth making.

Listing the sample space

When the stages are small, do not reason at all — draw the grid.

Two six-sided dice give a 6 by 6 table of 36 equally likely cells. Write the total in each cell and any question about totals becomes counting: 6 cells give a total of 7, so P(7) = 6/36 = 1/6; only 1 cell gives 2, so P(2) = 1/36.

That table also settles a misconception that reasoning alone rarely dislodges. There are 11 possible totals, from 2 to 12, but they are not equally likely, so P(7) is nowhere near 1/11. The cells are equally likely; the totals are not. Count cells.

For three coins a tree diagram does the same job: 8 branches, 8 equally likely outcomes, and P(exactly two heads) = 3/8 because three branches qualify.

At least one

"At least one gold in three draws" covers one gold, two golds and three golds. Listing them is slow and easy to get wrong. There is exactly one thing it rules out: no gold at all.

P(at least one) = 1 - P(none)

With 4 gold and 6 plain, drawing 2 without replacement:

P(no gold) = 6/10 x 5/9 = 30/90 = 1/3, so P(at least one gold) = 1 - 1/3 = 2/3.

Two checks are worth making every time. The answer must sit between 0 and 1. And "at least one" over several draws must be larger than the chance of gold on a single draw, because you are getting more chances at it.

Worked examples

Example 1

A pouch holds 5 red and 3 blue beads. Two are drawn without replacement. What is P(both red)?

  1. There are 8 beads in total, so P(first is red) = 5/8.
  2. The first bead is not put back. If it was red, 4 red beads remain out of 7 beads in all.
  3. So P(second is red, given the first was) = 4/7.
  4. P(both red) = 5/8 x 4/7 = 20/56 = 5/14.
  5. Check the direction: with replacement it would be 5/8 x 5/8 = 25/64, which is about 0.39, and 5/14 is about 0.36 — smaller, as it must be.

Example 2

Two fair six-sided dice are rolled. What is the probability that the total is more than 9?

  1. Lay out the sample space: a 6 by 6 grid, 36 equally likely cells.
  2. Totals above 9 are 10, 11 and 12.
  3. Count the cells: 10 comes from (4,6), (5,5), (6,4) — three cells. 11 comes from (5,6), (6,5) — two. 12 comes from (6,6) — one.
  4. That is 3 + 2 + 1 = 6 favourable cells out of 36.
  5. P(total above 9) = 6/36 = 1/6. Note that it is not 3/11, because the eleven possible totals are not equally likely.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

A jar holds 5 green and 5 silver counters. One is drawn, its colour noted, and PUT BACK. Then a second is drawn. What is the probability that both draws are green? Give a fraction in lowest terms.

Answer: 1/4

  1. P(first is green) = 5/10
  2. The token is replaced, so P(second is green) = 5/10 as well.
  3. P(both) = 5/10 x 5/10 = 25/100
  4. In lowest terms that is 1/4 (about 0.25).

Problem 2

Difficulty 3 of 5

A jar holds 7 silver, 4 green and 5 blue reed discs. Two are drawn one after the other and NOT put back. What is the probability that both are silver? Give a fraction in lowest terms.

Answer: 7/40

  1. P(first is silver) = 7/16
  2. After that draw 6 silver tokens remain out of 15, so P(second is silver | first was) = 6/15.
  3. P(both) = 7/16 x 6/15 = 42/240
  4. In lowest terms that is 7/40 (about 0.175).

Problem 3

Difficulty 4 of 5

Two fair 6-sided dice are rolled and their scores are added. The table shows every possible total. What is the probability of an even total? Give a fraction in lowest terms.

Answer: 1/2

  1. Every cell is equally likely, and there are 6 x 6 = 36 of them.
  2. Cells giving an even total: 18.
  3. P(an even total) = 18/36 = 1/2

Common mistakes

  • Using the same probability for the second draw when nothing was put back — counting with replacement in a problem without it.
  • Reducing the numerator but leaving the denominator unchanged after a draw, so the token half-leaves the jar.
  • Treating the eleven totals of two dice as equally likely and answering 1/11 for P(7).
  • Adding the stage probabilities instead of multiplying, which makes "both" more likely than either alone.
  • Answering the "at least one" question with the probability of none, or with the probability for a single draw.

What you should be able to do

  • Find the probability of a two-stage event with replacement.
  • Adjust both the numerator and the denominator when there is no replacement.
  • Read a probability off an organised list or a two-way table of a sample space.
  • Handle an "at least one" compound event through its complement.

Where this fits in the curriculum

Common Core

  • 7.SP.C.8

    Grade 7 — Find probabilities of compound events using organised lists, tables, tree diagrams and simulation.

  • 7.SP.C.8.A

    Grade 7 — Understand that the probability of a compound event is the fraction of outcomes in the sample space for which the event occurs.

  • HSS-CP.A.2

    High school — Understand independence, and that the probability of two independent events both occurring is the product of their probabilities.

Ontario

  • G8.D2

    Grade 8 — Describe the likelihood that events will happen, and use that information to make predictions.

    Ontario numbers its Data expectations per grade document and the specific numbering could not be verified line by line here, so this tag names the strand's OVERALL expectation rather than guessing at a specific one.

SAT

  • Problem Solving and Data Analysis

    Probabilities of compound events, almost always presented as a two-way table.

    The SAT rarely asks for a multi-stage probability in the abstract; it asks for one read out of a table of counts, which is the same computation in another dress.

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