A probability tree is useful only when its branches describe the process correctly. Multiplying neatly written fractions will not repair a second-stage probability copied from the wrong situation. Before calculating, identify what has happened, what remains available and which event the question asks for. Those decisions connect the representation to the actual experiment described in the task.
This original practice session supports the Hala O and A Level study guide and the Pakistan O Level revision plan. It is not an official examination paper or a claim about a local class. The decorative images are not probability diagrams or data sources. Use the quantities and conditions stated in each exercise.
Define the outcomes before assigning probabilities
Begin with an invented bag containing three red counters and two blue counters. Suppose one counter is chosen at random and every individual counter is equally likely to be selected. There are five equally likely individual outcomes, three of which are red. The probability of selecting red is therefore 3/5, and the probability of blue is 2/5.
Two colour names do not make the colour events equally likely. The colour categories contain different numbers of equally likely counters. A student who writes 1/2 because there are two colours has counted labels rather than the individual outcomes that the random selection makes equally likely. Explain that distinction before introducing a second selection.
Check that the probabilities of the mutually exclusive and exhaustive colour outcomes sum to one: 3/5 + 2/5 = 1. This is a useful consistency check. It does not by itself prove that every branch is correct, since two incorrect probabilities can also sum to one. The count and equal-likelihood assumption must still justify the values.
Describe replacement as part of the process
For the first two-stage version, return the selected counter to the bag and mix before making the second random selection under the same conditions. The bag again contains three red and two blue counters. The second selection therefore has the same colour probabilities as the first, regardless of which colour was selected earlier.
The probability of two reds is 3/5 multiplied by 3/5, which equals 9/25. The probability of red followed by blue is 3/5 multiplied by 2/5, which equals 6/25. Each product follows one ordered path through the process. Keep the order visible instead of listing two fractions without explaining which selections they represent.
A tree can show the two first-stage branches and the two second-stage branches from each. Every branch pair leaving a particular point should describe the outcomes available at that stage. Label the branch probabilities from the conditions, not from a remembered picture. A correctly drawn shape with incorrect values remains an incorrect model of the experiment.
Update the bag when the counter is not returned
Now change the process so that the first selected counter is kept out of the bag. The first-stage probabilities remain 3/5 and 2/5 because the starting bag has not changed. The second-stage probabilities depend on the first outcome. There are four counters left, but their colour composition differs between the first-red and first-blue branches.
After a red first selection, two red and two blue counters remain. The second red probability is 2/4 and the second blue probability is 2/4. After a blue first selection, three red and one blue remain. The corresponding probabilities are 3/4 and 1/4. Derive these values by describing the remaining bag on each branch.
The probability of two reds is now 3/5 multiplied by 2/4, which equals 3/10. This differs from 9/25 with replacement. Do not select the value that seems more familiar. Explain why removing a red counter changes the composition before the second selection. The process, not the appearance of the fractions, determines the calculation.
Distinguish an ordered event from an unordered event
In the no-replacement version, “red then blue” is one ordered event. Its probability is 3/5 multiplied by 2/4, which equals 3/10. “One red and one blue in either order” includes both red-then-blue and blue-then-red. The second path has probability 2/5 multiplied by 3/4, also 3/10.
The two ordered paths are mutually exclusive within this two-selection experiment: the first selection cannot be both red and blue. Add their probabilities to obtain 3/5 for one counter of each colour in either order. The addition is justified by the event being the union of these separate possibilities, not simply because the question contains the word “or.”
Read the requested event precisely. “First red and second blue,” “different colours” and “at least one red” refer to different collections of outcomes. A tree contains the possible paths, but the student must still decide which paths belong to the event. Highlight the relevant outcomes before adding their probabilities.
Check all the no-replacement paths together
For the same bag, the probability of two blues is 2/5 multiplied by 1/4, which equals 1/10. The probabilities of red-red, red-blue, blue-red and blue-blue are therefore 3/10, 3/10, 3/10 and 1/10. They sum to one, as they should for these exhaustive mutually exclusive ordered colour outcomes.
This total is another consistency check, not a replacement for deriving each branch. If the total is wrong, inspect the remaining counts and the multiplication on each path. If the total is right, still inspect whether the branches match the stated process. An error that redistributes probability between paths might leave the overall total unchanged.
Ask the student to explain why the red-blue and blue-red probabilities happen to agree here. Their second-stage fractions are not the same, but the products are equal. This observation should not be turned into an unrestricted rule about every multi-stage experiment. Use the actual structure and probabilities when comparing other scenarios.
Use a complement when it matches the event
In the no-replacement bag example, “at least one red” includes red-red, red-blue and blue-red. The only excluded colour outcome is blue-blue, whose probability is 1/10. Therefore the probability of at least one red is 1 − 1/10 = 9/10. The complement route is short because the excluded event is simple to describe and calculate.
Check the meaning of “at least one.” It includes two reds as well as exactly one red. The probability of exactly one red is 3/5 from the different-colours calculation, so it is not the same event. A student who uses these phrases interchangeably needs an outcome-selection repair before another multiplication exercise.
For a fresh check, ask for no red in two selections. That event is two blues and has probability 1/10 in this no-replacement model. Comparing no red, exactly one red and at least one red helps the student connect language to event sets. Keep the experiment fixed while changing the event so the source of the difference remains visible.
Change the starting bag to test transfer
Use a new original bag containing four green and three yellow counters, with two random selections without replacement. Ask for two greens. The first probability is 4/7. After a green is removed, three greens remain among six counters, so the second probability is 3/6. Multiplying gives 12/42, or 2/7.
Ask for the reason behind each number. Seven is the initial total; six is the remaining total; three is the green count after the specified first outcome. If the student copies 2/4 from the earlier bag, the procedure has not yet transferred. Return to the remaining-outcomes description rather than asking them to memorise another completed tree.
The changed-condition guide gives older students a related routine for finding the step affected by a new assumption. For this O Level exercise, keep the notation and required detail appropriate to the student's course. The general habit is to recheck conditions before reusing a familiar calculation.
Choose the representation for the decision it supports
A tree is useful for displaying stages and conditional branches. An outcome table can help when the outcomes form a clear grid and their probabilities are understood. Do not assume that every cell of a table is equally likely merely because the cells look the same size. The experiment must justify any equal-probability counting method.
Whichever representation is used, write the event in words and identify its outcomes. The general probability trees and tables guide links this session to the subject cluster. The Hala practice-selection guide explains how a family can use the attempt to choose the next support question.
Finish with an independent attempt after the model is closed. Keep any meaningful hint visible and bring a precise unresolved question to the teacher. For teaching availability and included work review, confirm details through the admissions FAQ. This article does not establish an individual marking service, class format or examination arrangement in Hala.
Frequently asked questions
Why is red not one half when there are two colours?
The equally likely individual outcomes are the counters, not the colour labels. With three red and two blue counters under the stated random-selection conditions, red has probability 3/5. Different category sizes can produce different probabilities.
Which probabilities change without replacement?
The first selection uses the original bag. The second selection uses the counters remaining after the first outcome on that branch. Both the total and the relevant colour count must be checked rather than copied from the starting bag.
When do I add two path probabilities?
Add when the requested event consists of those mutually exclusive alternatives. For one red and one blue in either order, include red-blue and blue-red. Explain which outcomes belong to the event rather than relying on a single keyword.
Is exactly one red the same as at least one red?
No. Exactly one excludes the two-red outcome, while at least one includes it. Identify the relevant paths before calculating. In the original no-replacement example, their probabilities are 3/5 and 9/10 respectively.
Does a total probability of one prove my tree is correct?
No. It is a useful consistency check, but incorrect branch values can sometimes still sum to one. Derive each probability from the stated process and remaining outcomes, then use the total as an additional check.
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