Math Problem Statement

Supply the missing quantities. Figure Description A tree diagram has two stages. Stage 1 has three nodes and stage 2 has six nodes. In stage 1, the branch from the starting position to node A is labeled 0.3. The branch from the starting position to node B is labeled 0.1. The branch from the starting position to node C is an answer blank. In stage 2, the branch from node A to node E is an answer blank. The branch from node A to node F is labeled 0.5. In stage 2, the branch from node B to node G is an answer blank. The branch from node B to node H is labeled 0.7. In stage 2, the branch from node C to node I is an answer blank. The branch from node C to node J is an answer blank. Node I is labeled P(I ∩ C) = 0.24. Node J is labeled P(J ∩ C) = 0.36. Outcome P(A ∩ E)

P(A ∩ F)

P(B ∩ G)

P(B ∩ H)

Solution

To solve this problem, we need to use the principles of probability, specifically the multiplication rule of probability. In a tree diagram, the probability of reaching a specific outcome is the product of the probabilities along the branches leading to that outcome. Let's work through the missing values step-by-step.

Stage 1:

  • The sum of all probabilities leading out of the starting point must be equal to 1.
    • P(A) = 0.3 (already given)
    • P(B) = 0.1 (already given)
    • Let P(C) be the missing probability. So we have: P(A)+P(B)+P(C)=1P(A) + P(B) + P(C) = 1 0.3+0.1+P(C)=10.3 + 0.1 + P(C) = 1 P(C)=10.4=0.6P(C) = 1 - 0.4 = 0.6 Therefore, P(C)=0.6P(C) = 0.6.

Stage 2:

  • From node A:

    • P(A ∩ E): Let this branch probability be P(EA)P(E \mid A). We know that the other branch (A → F) has a probability of 0.5, so: P(EA)+P(FA)=1P(E \mid A) + P(F \mid A) = 1 P(EA)+0.5=1P(E \mid A) + 0.5 = 1 P(EA)=0.5P(E \mid A) = 0.5
    • Now, to get the joint probabilities: P(AE)=P(A)×P(EA)=0.3×0.5=0.15P(A ∩ E) = P(A) \times P(E \mid A) = 0.3 \times 0.5 = 0.15 P(AF)=P(A)×P(FA)=0.3×0.5=0.15P(A ∩ F) = P(A) \times P(F \mid A) = 0.3 \times 0.5 = 0.15
  • From node B:

    • P(B ∩ G): Let P(GB)P(G \mid B) be the branch probability for B → G. Since B → H is 0.7: P(GB)+P(HB)=1P(G \mid B) + P(H \mid B) = 1 P(GB)+0.7=1P(G \mid B) + 0.7 = 1 P(GB)=0.3P(G \mid B) = 0.3
    • Now, to get the joint probabilities: P(BG)=P(B)×P(GB)=0.1×0.3=0.03P(B ∩ G) = P(B) \times P(G \mid B) = 0.1 \times 0.3 = 0.03 P(BH)=P(B)×P(HB)=0.1×0.7=0.07P(B ∩ H) = P(B) \times P(H \mid B) = 0.1 \times 0.7 = 0.07
  • From node C:

    • We are already given the joint probabilities for outcomes I and J: P(IC)=0.24,P(JC)=0.36P(I ∩ C) = 0.24, \quad P(J ∩ C) = 0.36
    • We can verify that these probabilities add up to the total probability of C: P(IC)+P(JC)=1P(I \mid C) + P(J \mid C) = 1 0.24+0.36=0.6(which matches P(C)=0.6)0.24 + 0.36 = 0.6 \quad \text{(which matches } P(C) = 0.6 \text{)}

Final Outcome:

  1. P(AE)=0.15P(A ∩ E) = 0.15
  2. P(AF)=0.15P(A ∩ F) = 0.15
  3. P(BG)=0.03P(B ∩ G) = 0.03
  4. P(BH)=0.07P(B ∩ H) = 0.07

These are the missing probabilities based on the given tree structure.

Do you want more details or have any questions?

Here are 5 questions related to this problem:

  1. How would the probabilities change if the probability of A was different?
  2. How does the multiplication rule of probability work in tree diagrams?
  3. Can you explain the difference between conditional and joint probabilities?
  4. How would you find the probability of reaching a particular node in a tree with more than two stages?
  5. What would happen if the probabilities from each node did not sum to 1?

Tip: Always check that probabilities at each branching point sum to 1; this ensures the diagram reflects a complete set of outcomes.

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Math Problem Analysis

Mathematical Concepts

Probability
Tree Diagrams
Joint Probability
Conditional Probability

Formulas

Sum of probabilities for branches from a node must equal 1: P(A) + P(B) + P(C) = 1
Multiplication rule for joint probability: P(A ∩ E) = P(A) × P(E | A)

Theorems

Multiplication Rule of Probability
Sum Rule of Probability

Suitable Grade Level

Grades 11-12