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Statistics and Probability - Venn diagrams and tree diagrams

Grade 10IB

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

πŸ”‘Concepts

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The Universal Set (ΞΎ\xi): Represented visually as a large bounding rectangle, the universal set contains all possible outcomes for a given scenario. Any outcomes not belonging to specific subsets are written inside this rectangle but outside the circles.

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Intersections (A∩BA \cap B): This is the region where two or more circles overlap in a Venn diagram. It represents the logical 'AND', meaning the elements belong to both set AA and set BB. If the circles do not overlap, the intersection is empty, indicating the events are mutually exclusive.

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Unions (AβˆͺBA \cup B): This concept encompasses all elements found in circle AA, circle BB, or both. Visually, it is the total area covered by both circles. It represents the logical 'OR'.

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Complements (Aβ€²A'): The complement of set AA consists of all outcomes in the universal set that are NOT in set AA. In a Venn diagram, this is everything shaded outside the boundary of circle AA but still within the rectangular universal set.

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Tree Diagram Structure: A visual tool used to map out probabilities for sequential or multiple events. It starts at a single point and branches out; each 'branch' represents a possible outcome. The probabilities for each set of branches originating from a single node must always sum to 1.

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Path Probabilities (Multiplication Rule): To find the probability of a combined outcome along a specific path in a tree diagram (e.g., getting a Head then a Tail), you multiply the probabilities written on the branches along that path. This represents P(A∩B)P(A \cap B).

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Combining Outcomes (Addition Rule): If a specific result can be achieved through multiple different paths in a tree diagram, you calculate the probability of each path separately and then add the results together. This is used for 'OR' scenarios across different sequences.

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Conditional Probability: In Venn diagrams, this is the probability of AA occurring given BB has already occurred (P(A∣B)P(A|B)), visually represented by focusing only on circle BB as the new universal set. In tree diagrams, the probabilities on the second set of branches are inherently conditional on the first event.

πŸ“Formulae

P(A)+P(Aβ€²)=1P(A) + P(A') = 1

P(AβˆͺB)=P(A)+P(B)βˆ’P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A∩B)=P(A)Γ—P(B)P(A \cap B) = P(A) \times P(B) (For independent events only)

P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

P(A∩B)=P(A)Γ—P(B∣A)P(A \cap B) = P(A) \times P(B|A) (General multiplication rule)

πŸ’‘Examples

Problem 1:

In a class of 30 students, 18 study Art, 15 study Music, and 8 study both. Find the probability that a student chosen at random studies neither Art nor Music.

Solution:

  1. Identify known values: n(ξ)=30n(\xi) = 30, n(A∩M)=8n(A \cap M) = 8.
  2. Find students who study only Art: 18βˆ’8=1018 - 8 = 10.
  3. Find students who study only Music: 15βˆ’8=715 - 8 = 7.
  4. Calculate students in the union (AβˆͺMA \cup M): 10+8+7=2510 + 8 + 7 = 25.
  5. Find students who study neither: 30βˆ’25=530 - 25 = 5.
  6. Probability calculation: P(Neither)=530=16P(\text{Neither}) = \frac{5}{30} = \frac{1}{6}.

Explanation:

We use a Venn diagram approach. First, we fill the intersection (8), then subtract that from the individual totals to find the unique parts of each circle. The remainder of the universal set represents the students who study neither subject.

Problem 2:

A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random without replacement. Find the probability that both marbles are the same color.

Solution:

  1. Draw a tree diagram. The first draw has probabilities P(R1)=58P(R1) = \frac{5}{8} and P(B1)=38P(B1) = \frac{3}{8}.
  2. Second draw probabilities (without replacement):
    • If Red first: P(R2∣R1)=47P(R2|R1) = \frac{4}{7} and P(B2∣R1)=37P(B2|R1) = \frac{3}{7}.
    • If Blue first: P(R2∣B1)=57P(R2|B1) = \frac{5}{7} and P(B2∣B1)=27P(B2|B1) = \frac{2}{7}.
  3. Identify paths for 'same color': (Red, Red) or (Blue, Blue).
  4. Path 1 (R, R): 58Γ—47=2056\frac{5}{8} \times \frac{4}{7} = \frac{20}{56}.
  5. Path 2 (B, B): 38Γ—27=656\frac{3}{8} \times \frac{2}{7} = \frac{6}{56}.
  6. Total Probability: 2056+656=2656=1328\frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}.

Explanation:

Since the marbles are not replaced, the denominator and numerator change for the second event. We multiply along the branches for each valid path and add the results because the two paths are mutually exclusive ways to get 'same color'.