QUESTION IMAGE
Question
- a financial planner has three portfolios: a, b, and c. because investors have different tolerances for risks, 20% of people are likely to invest in portfolio a, 30% are likely to invest in b, and 50% are likely to invest in c. each portfolio has both stocks and bonds, and investors are equally likely to choose either. this is a tree diagram that represents the probability of investors choosing the different financial products. tree diagram with probabilities 0.2, 0.3, z, and 0.5 for each subsequent branch what is the value of z? options: 0.5, 0.4, 0.3, 0.6
Step1: Recall total probability rule
The sum of probabilities of all mutually exclusive and exhaustive events is 1. Here, investing in A, B, or C are exhaustive and mutually exclusive. So, \( P(A) + P(B) + P(C) = 1 \).
Step2: Substitute known probabilities
We know \( P(A) = 0.2 \), \( P(B) = 0.3 \), and \( P(C)=Z \). So, \( 0.2 + 0.3 + Z = 1 \).
Step3: Solve for Z
Simplify the equation: \( 0.5 + Z = 1 \). Subtract 0.5 from both sides: \( Z = 1 - 0.5 = 0.5 \)? Wait, no, wait. Wait, 0.2 + 0.3 is 0.5, so 1 - 0.5 is 0.5? Wait, no, wait the options have 0.5? Wait, no, wait the problem says 20% (0.2) for A, 30% (0.3) for B, so C should be 1 - 0.2 - 0.3 = 0.5? Wait, but let's check again. The tree diagram's first level is the probability of choosing each portfolio. So the sum of the first - level probabilities (0.2, 0.3, Z) should be 1. So \( 0.2+0.3 + Z = 1 \). Then \( Z = 1-(0.2 + 0.3)=1 - 0.5 = 0.5 \). Wait, but the options include 0.5. Wait, maybe I made a mistake? Wait, no, the total probability of all possible first - stage events (choosing A, B, or C) must sum to 1. So 0.2 (A) + 0.3 (B)+Z (C)=1. So Z = 1 - 0.2 - 0.3 = 0.5.
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