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Statistics and Probability - Sample Spaces and Tree Diagrams

Grade 8IB

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

🔑Concepts

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A Sample Space (denoted by SS) is the set of all possible outcomes of a random experiment. For example, for a coin toss, S={H,T}S = \{H, T\}.

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An Event is a subset of the sample space consisting of one or more outcomes. We denote the number of outcomes in an event as n(E)n(E) and in the sample space as n(S)n(S).

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A Tree Diagram is a visual representation used to list all possible outcomes of a sequence of events. Each 'branch' represents a possible choice or outcome.

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The Fundamental Counting Principle states that if there are mm ways to perform one task and nn ways to perform another, there are m×nm \times n total ways to perform both tasks.

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For multi-stage events, the probability of a specific outcome is found by multiplying the probabilities along the branches of the tree diagram.

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The sum of probabilities of all possible outcomes in a sample space must always equal 11.

📐Formulae

P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

P(A′)=1−P(A)P(A') = 1 - P(A) (Complementary Events)

Total Outcomes=n1×n2×⋯×nk\text{Total Outcomes} = n_1 \times n_2 \times \dots \times n_k

P(A then B)=P(A)×P(B)P(A \text{ then } B) = P(A) \times P(B) (for independent events)

💡Examples

Problem 1:

A fair coin is tossed twice. List the sample space using a tree diagram and find the probability of getting exactly one Head.

Solution:

The tree diagram has two stages:

  1. First toss: HH or TT
  2. Second toss: HH or TT for each result of the first toss.

The sample space is S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}. Total outcomes n(S)=4n(S) = 4. Favorable outcomes for 'exactly one Head' are E={HT,TH}E = \{HT, TH\}, so n(E)=2n(E) = 2. P(exactly one Head)=24=12P(\text{exactly one Head}) = \frac{2}{4} = \frac{1}{2}

Explanation:

We list all possible paths in the tree diagram to find n(S)n(S) and then count how many of those paths satisfy the specific condition.

Problem 2:

A spinner has 3 equal sections colored Red (RR), Blue (BB), and Green (GG). If you spin it once and then roll a standard six-sided die, how many total outcomes are in the sample space?

Solution:

Number of outcomes for the spinner: n1=3n_1 = 3 Number of outcomes for the die: n2=6n_2 = 6 Total outcomes using the Fundamental Counting Principle: Total=3×6=18\text{Total} = 3 \times 6 = 18

Explanation:

To find the total number of outcomes for combined events, we multiply the number of options for each individual event.

Problem 3:

A bag contains 3 red balls and 2 blue balls. A ball is drawn, its color is noted, and it is replaced before a second ball is drawn. Draw a tree diagram to find the probability of drawing two red balls.

Solution:

Probability of Red (RR) = 35\frac{3}{5}. Probability of Blue (BB) = 25\frac{2}{5}. Since the ball is replaced, the probabilities remain the same for the second draw. P(R,R)=P(R)×P(R)P(R, R) = P(R) \times P(R) P(R,R)=35×35=925P(R, R) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} In decimal form: 0.360.36.

Explanation:

Because the event is 'with replacement', the events are independent. We multiply the probabilities of the 'Red' branches for both stages.