Mathematics · 17 / 23
Probability: read a tree
High school · intermediate · Worked example and 4 exercises with solutions, at your pace.
The principle, simply
A probability lies between zero and one. Branches from the same node show conditional possibilities adding to one. Multiply along a path for an intersection; add mutually exclusive paths for an event. Independence requires a reason, not an assumption.
Multiply along a path; add disjoint paths for one event. This partial tree omits branches after not A.
- 01Node totals one
- 02Multiply a path
- 03Add exclusive paths
- 04Check zero to one
WORKED EXAMPLE
From principle to reasoning
P(A) = 0.4 and P(B given A) = 0.5 give a joint path probability of 0.2. An unconditional P(B) cannot automatically replace P(B given A).
Reconstruct total probability
If P(A)=0.4, P(B|A)=0.5 and P(B|not A)=0.1, disjoint paths give P(B)=0.4×0.5+0.6×0.1=0.26. Conversely P(A|B)=0.20/0.26≈0.769: reversing the condition changes the question.
Solve a problem in this subject
- Define the objects
State unknowns, domain and data. Real square roots need nonnegative arguments; denominators must be nonzero.
- Choose a transformation
Name the rule: expansion, equivalent equations, factoring, differentiation. Check its conditions. Squaring an equation requires testing candidates in the original.
- Show the reasoning
Each line should follow from the previous one. Keep exact values where possible and distinguish equality, approximation and implication.
- Test and conclude
Substitute into the original expression. Check graph points and domain boundaries. State a solution set or the required unit.
YOUR TURN
01 · Try before checking
Two independent fair coin tosses: probability of two heads?
One hint
Each heads branch has probability 1/2.
Reveal and understand the solution
(1/2) × (1/2) = 1/4. HH, HT, TH and TT are equally likely.
Compare method, conditions and answer; identify the exact part to revisit.
IMPROVE THROUGH PRACTICE
Three additional challenges with solutions
Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.
A fair six-sided die is rolled. Find the probability of an even number.
A hint to get started
Count favourable equally likely outcomes.
Step-by-step solution
- Even outcomes are 2, 4 and 6.
- There are three out of six.
- P(even) = 1/2.
Which step did you miss? Retry without help, then explain why the method works.
P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A ∪ B).
A hint to get started
Adding directly counts the intersection twice.
Step-by-step solution
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- 0.4 + 0.5 − 0.2 = 0.7.
- This formula does not require independence.
Which step did you miss? Retry without help, then explain why the method works.
Draw twice without replacement from 3 red and 2 blue balls. Find the probability of two reds.
A hint to get started
After one red, both the total and red count change.
Step-by-step solution
- The first red has probability 3/5.
- Then 2 reds remain among 4 balls: 2/4.
- Multiply: 3/5 × 2/4 = 3/10.
Which step did you miss? Retry without help, then explain why the method works.
The trap to avoid
Do not add along a path or reverse conditional probabilities.
Without notes, can you…
- Explain the principle in your own words?
- Rebuild the diagram and its connections?
- Solve the exercise and justify the method?
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