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Statistics and Probability - Tree diagrams

Grade 12IB_AA

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

🔑Concepts

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A tree diagram is a visual tool used to represent the outcomes of a multi-stage experiment and their associated probabilities.

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The sum of the probabilities on the branches originating from any single node must always equal 1: ∑P=1\sum P = 1.

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To find the probability of a specific sequence of events (the 'AND' rule), multiply the probabilities along the branches of that path: P(A∩B)=P(A)×P(B∣A)P(A \cap B) = P(A) \times P(B|A).

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To find the probability of an event that can occur through multiple different paths (the 'OR' rule), add the probabilities of those individual paths together.

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Conditional probability is naturally represented in tree diagrams. The second set of branches represents the probability of an event occurring given that the first event has already occurred, denoted as P(B∣A)P(B|A).

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Sampling 'with replacement' means probabilities remain constant for each stage. Sampling 'without replacement' means the probabilities change at each stage because the total number of items decreases.

📐Formulae

P(A∩B)=P(A)×P(B∣A)P(A \cap B) = P(A) \times P(B|A) pieces

P(B)=P(A∩B)+P(A′∩B)P(B) = P(A \cap B) + P(A' \cap B)

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

∑P(branches from one node)=1\sum P(\text{branches from one node}) = 1

💡Examples

Problem 1:

A bag contains 5 red balls and 3 green balls. Two balls are drawn at random one after another without replacement. Find the probability that both balls are of the same color.

Solution:

  1. Let RR be the event of picking a red ball and GG be the event of picking a green ball.
  2. First draw: P(R1)=58P(R_1) = \frac{5}{8} and P(G1)=38P(G_1) = \frac{3}{8}.
  3. Second draw (without replacement):
    • If the first was Red: P(R2∣R1)=47P(R_2|R_1) = \frac{4}{7} and P(G2∣R1)=37P(G_2|R_1) = \frac{3}{7}.
    • If the first was Green: P(R2∣G1)=57P(R_2|G_1) = \frac{5}{7} and P(G2∣G1)=27P(G_2|G_1) = \frac{2}{7}.
  4. We want P(Same Color)=P(R1∩R2)+P(G1∩G2)P(\text{Same Color}) = P(R_1 \cap R_2) + P(G_1 \cap G_2).
  5. P(R1∩R2)=58×47=2056P(R_1 \cap R_2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}.
  6. P(G1∩G2)=38×27=656P(G_1 \cap G_2) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}.
  7. Total probability: 2056+656=2656=1328\frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}.

Explanation:

Since the sampling is without replacement, the denominator and the numerator of the specific color chosen decrease by 1 for the second branch. We identify the two paths that satisfy 'same color' (RRRR and GGGG), calculate their path probabilities by multiplying, and then add them.

Problem 2:

In a certain population, 1%1\% of people have a specific disease. A diagnostic test is 95%95\% accurate (it gives a positive result for 95%95\% of people with the disease and a negative result for 95%95\% of people without it). Find the probability that a person has the disease given they tested positive.

Solution:

  1. Let DD be having the disease and T+T+ be a positive test result.
  2. P(D)=0.01P(D) = 0.01, P(D′)=0.99P(D') = 0.99.
  3. P(T+∣D)=0.95P(T+|D) = 0.95 (True Positive) and P(T+∣D′)=0.05P(T+|D') = 0.05 (False Positive).
  4. Total probability of testing positive P(T+)P(T+): P(T+)=P(D∩T+)+P(D′∩T+)P(T+) = P(D \cap T+) + P(D' \cap T+) P(T+)=(0.01×0.95)+(0.99×0.05)=0.0095+0.0495=0.059P(T+) = (0.01 \times 0.95) + (0.99 \times 0.05) = 0.0095 + 0.0495 = 0.059
  5. Use Bayes' Theorem: P(D∣T+)=P(D∩T+)P(T+)P(D|T+) = \frac{P(D \cap T+)}{P(T+)} P(D∣T+)=0.00950.059≈0.161P(D|T+) = \frac{0.0095}{0.059} \approx 0.161.

Explanation:

This example uses a tree diagram to organize the conditional probabilities. To find the probability of the disease given a positive result, we divide the probability of the 'Disease and Positive' path by the sum of all paths leading to a 'Positive' result.