Find and Fix Errors in Compound Probability | Grade 7 Math | LevelKite

You already know how to find the probability of a compound event, an event that combines the results of two or more chance stages, such as tossing heads on a coin and then rolling a 5 on a number cube (a cube with faces numbered 1 to 6). In this lesson you switch roles. You read someone else's solution, find the step where it goes wrong, explain why that step fails, and fix it. This is called error analysis.

Two words will come up often. An outcome is one possible result of the whole activity, such as heads together with a 5. The sample space is the list of all the possible outcomes.

Start with why counting can give a probability at all. Think of the sample space as a set of equal shares of chance. Suppose a sample space has 4 outcomes and every outcome is equally likely, meaning each one has the same chance of happening. Then an event that happens in 2 of the outcomes holds 2 of the 4 equal shares, so its probability is $\frac{2}{4}$ . A coin or number cube is fair when all of its results are equally likely.

For a compound event, the sample space is made of ordered outcomes. An ordered outcome records the result of each stage in order, so heads then tails is a different outcome from tails then heads.

If a solution's list does not match the real situation, its count can give a wrong probability. Error analysis means finding the place where the list or the count stopped matching the situation.

Key concept

When favourable over total can be trusted

A favourable outcome is an outcome in which the event happens. The probability of a compound event is the fraction of the outcomes in the sample space that are favourable, as long as both of these conditions hold:

the sample space lists every ordered outcome exactly once, and
all of those outcomes are equally likely.

When you draw items without replacement (you take an item out and do not put it back), fewer items are available for the next draw, so the next draw has a smaller total. That changes the later denominators. Draws like these are called dependent events, because the first draw changes which outcomes are available for the second.

\text{probability of the event} = \frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

To check a solution, make three moves:

Find the first step that breaks one of the checks in the table below.
Explain why that step fails.
Fix the step and carry the corrected work through to a final answer.

Three checks for a compound probability solution

Check	Question to ask	What an error looks like
Ordered and complete	Does the sample space list every ordered outcome exactly once?	The list merges two different orders into one outcome, leaves an outcome out, or shows one twice.
Equally likely	Does every outcome in the list have the same chance?	The solution counts outcomes with different chances as equal shares.
Replacement	Does a drawn item go back? If it stays out, does the next draw use the smaller number left?	The item stays out, but the solution still uses the starting number of items for the next draw.

Worked example

Two orders counted as one

Ravi tosses a fair coin twice, so heads (H) and tails (T) are equally likely on each toss. He lists 3 outcomes: two heads, two tails, and one of each. He says the probability of getting one head and one tail is $\frac{1}{3}$ . Find Ravi's error and correct it.

Start with the first check by writing the ordered outcomes yourself, first toss then second toss, so HT means heads then tails. The first toss has 2 results, and each of them can be followed by 2 results on the second toss, so the list has 2 times 2 outcomes.
$2 \times 2 = 4: \quad \text{HH},\ \text{HT},\ \text{TH},\ \text{TT}$
Find the error by comparing lists. Ravi's third outcome, one of each, stands for both HT and TH (tails then heads). Those are two different ordered outcomes, so his sample space does not list every ordered outcome exactly once.
Explain why that gives a wrong answer. The coin is fair and one toss does not affect the other, so the 4 ordered outcomes are equally likely. Ravi's 3 results are not: one of each happens in 2 of the 4 outcomes, but two heads happens in only 1. Favourable over total needs equally likely outcomes, so dividing by 3 is not valid.
Fix it with the complete list. The favourable outcomes, with one head and one tail, are HT and TH, so 2 of the 4 equally likely outcomes are favourable. Dividing the numerator and the denominator by the same number, 2, keeps the value of the fraction the same and makes it simpler.
$\frac{2}{4} = \frac{2 \div 2}{4 \div 2} = \frac{1}{2}$

Answer

Ravi counted the two ordered outcomes HT and TH as one outcome. With all 4 equally likely ordered outcomes listed, the probability of one head and one tail is $\frac{2}{4} = \frac{1}{2}$ , not $\frac{1}{3}$ .

Worked example

Results that are not equally likely

A spinner has 4 equal sections, so each section is equally likely. Two sections are red, one is blue and one is green. Lena spins it twice. She says each spin has 3 results (red, blue or green), so there are $3 \times 3 = 9$ outcomes and the probability of red on both spins is $\frac{1}{9}$ . Find Lena's error and correct it.

Lena's count of 9 covers every ordered pair of her 3 results once, so the first check finds nothing wrong. Move to the equally likely check. The 4 sections are equally likely, and red covers 2 of them while blue covers 1 and green covers 1. Red is twice as likely as blue or as green, so her 3 results are not equally likely and her 9 pairs are not equal shares.
Fix the sample space by using the sections, because they are equally likely. Name the red sections R1 and R2, the blue section B and the green section G. The first spin does not affect the second, so each of the 4 sections on the first spin can be followed by any of the 4 sections on the second.
$4 \times 4 = 16 \text{ ordered outcomes}$
Count the favourable outcomes, which have a red section on both spins. The first spin can land on R1 or R2, and each of those can be followed by R1 or R2 on the second spin.
$2 \times 2 = 4: \quad (\text{R1}, \text{R1}),\ (\text{R1}, \text{R2}),\ (\text{R2}, \text{R1}),\ (\text{R2}, \text{R2})$
The 16 ordered outcomes are each counted once and are equally likely, so favourable over total applies. Dividing the numerator and the denominator by 4 keeps the value of the fraction the same.
$\frac{4}{16} = \frac{4 \div 4}{16 \div 4} = \frac{1}{4}$

Answer

Lena treated red, blue and green as equally likely, but red covers 2 of the 4 equal sections. With the 16 equally likely ordered outcomes, the probability of red on both spins is $\frac{4}{16} = \frac{1}{4}$ , not $\frac{1}{9}$ .

Worked example

A marble that stays out

A bag holds 5 marbles of the same size: 3 red and 2 blue. Without looking, Omar draws one marble, keeps it out of the bag, and then draws a second marble. He says there are $5 \times 5 = 25$ outcomes and $3 \times 3 = 9$ of them are red then red, so the probability of two red marbles is $\frac{9}{25}$ . Find Omar's error and correct it.

Find the error with the replacement check. Omar keeps the first marble out, so the draws are dependent events and only 4 marbles are in the bag for the second draw. He used 5 for the second draw, which is right only when the first marble goes back.
Fix the total. Name the red marbles R1, R2 and R3 and the blue marbles B1 and B2. Each of the 5 possible first marbles can be followed by any of the 4 marbles still in the bag. At each draw every marble in the bag has the same chance, so these ordered outcomes are equally likely.
$5 \times 4 = 20 \text{ ordered outcomes}$
Fix the favourable count, which has the same problem. The first marble can be any of the 3 red ones. That marble is now out of the bag, so the second can be either of the 2 red marbles that are left, not 3.
$3 \times 2 = 6 \text{ favourable outcomes}$
The 20 ordered outcomes are each counted once and are equally likely, so write favourable over total. Dividing the numerator and the denominator by 2 keeps the value of the fraction the same.
$\frac{6}{20} = \frac{6 \div 2}{20 \div 2} = \frac{3}{10}$

Answer

Omar counted as if the first marble went back into the bag. Without replacement there are 20 equally likely ordered outcomes and 6 of them are red then red, so the probability of two red marbles is $\frac{6}{20} = \frac{3}{10}$ , not $\frac{9}{25}$ .

Takeaway. A compound probability is only as good as its sample space. Before you trust favourable over total, check that the sample space lists every ordered outcome exactly once and that those outcomes are equally likely. When someone draws an item and does not put it back, check that the next draw uses the smaller number of items that are left. If a check fails, you have found the error: name the step, explain why it fails, and rebuild the count from a correct sample space.