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Statistics and Probability - Probability of combined events using tree diagrams and Venn diagrams

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Venn Diagrams: A visual tool used to represent sets and their relationships within a universal set, denoted as ξ\xi. It is drawn as a large rectangle (the universal set) containing overlapping circles (subsets). The overlapping region of two circles AA and BB represents the intersection A∩BA \cap B (outcomes in both), while the entire area covered by both circles represents the union A∪BA \cup B (outcomes in AA, BB, or both).

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Intersection and Union: The intersection (A∩BA \cap B) refers to elements common to both sets, appearing in the shared 'football-shaped' region of a Venn diagram. The union (A∪BA \cup B) refers to all elements in either set or both, covering the combined area of the two circles. If circles do not overlap, they are mutually exclusive, meaning P(A∩B)=0P(A \cap B) = 0.

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Complementary Events: The complement of an event AA, denoted as A′A', represents all outcomes in the sample space that are NOT in AA. Visually, if AA is a circle in a Venn diagram, A′A' is the entire area inside the rectangle but outside the circle. The total probability P(A)+P(A′)=1P(A) + P(A') = 1.

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Tree Diagrams for Combined Events: A branching diagram used to map out all possible outcomes of two or more successive events. Each set of branches stems from a single node, and the probabilities on the branches from any single point must sum to 11. To find the probability of a specific path (e.g., event XX then event YY), you multiply the probabilities along the branches.

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Independent vs. Dependent Events: Events are independent if the outcome of the first does not change the probability of the second. In a tree diagram for independent events (like rolling a die twice), the probabilities on the second set of branches are identical to the first. In dependent events (like drawing cards without replacement), the probabilities change on the second set of branches because the total number of outcomes decreases.

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Calculating Probabilities from Trees: To find the probability of a combined outcome, such as 'at least one' or 'different results', you identify all valid paths on the tree diagram, calculate the product of each path, and then add those products together.

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Sample Space and ξ\xi: The sample space is the set of all possible outcomes of an experiment. In Venn diagrams, the universal set symbol ξ\xi or UU is placed in the corner of the rectangle to indicate every possible outcome being considered, including those that do not fall into specific categories AA or BB (placed outside the circles).

📐Formulae

P(A)=number of outcomes in Atotal number of outcomes in ξP(A) = \frac{\text{number of outcomes in } A}{\text{total number of outcomes in } \xi}

P(A′)=1−P(A)P(A') = 1 - P(A)

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) (only if events are independent)

∑P(all outcomes)=1\sum P(\text{all outcomes}) = 1

💡Examples

Problem 1:

In a class of 30 students, 15 play soccer (SS), 12 play basketball (BB), and 5 play both. If a student is chosen at random, find the probability that the student plays soccer but NOT basketball.

Solution:

  1. Identify the number of students in the intersection: n(S∩B)=5n(S \cap B) = 5.
  2. Calculate students who play only soccer: n(S only)=n(S)−n(S∩B)=15−5=10n(S \text{ only}) = n(S) - n(S \cap B) = 15 - 5 = 10.
  3. Calculate students who play only basketball: n(B only)=n(B)−n(S∩B)=12−5=7n(B \text{ only}) = n(B) - n(S \cap B) = 12 - 5 = 7.
  4. Find the total number of students who play at least one sport: 10+5+7=2210 + 5 + 7 = 22.
  5. The number of students playing soccer but not basketball is 1010.
  6. The probability is P(S∩B′)=1030=13P(S \cap B') = \frac{10}{30} = \frac{1}{3}.

Explanation:

We use a Venn diagram approach. By subtracting the intersection from the total soccer players, we isolate those who are exclusively in the soccer set. The probability is then that count divided by the total universal set of students.

Problem 2:

A bag contains 4 red marbles and 6 blue marbles. Two marbles are drawn one after the other without replacement. Find the probability of drawing one marble of each color.

Solution:

  1. Set up the tree diagram. First draw: P(R)=410P(R) = \frac{4}{10}, P(B)=610P(B) = \frac{6}{10}.
  2. Second draw if Red was first: P(R∣R)=39P(R|R) = \frac{3}{9}, P(B∣R)=69P(B|R) = \frac{6}{9}.
  3. Second draw if Blue was first: P(R∣B)=49P(R|B) = \frac{4}{9}, P(B∣B)=59P(B|B) = \frac{5}{9}.
  4. Identify paths for 'one of each color': (Red, then Blue) or (Blue, then Red).
  5. P(R then B)=410×69=2490P(R \text{ then } B) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}.
  6. P(B then R)=610×49=2490P(B \text{ then } R) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}.
  7. Total probability = 2490+2490=4890=815\frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}.

Explanation:

Since the marbles are not replaced, the denominator decreases for the second event. We multiply along the branches for each valid scenario and then add the resulting probabilities because either scenario satisfies the condition.