Math Problem Statement
Three boys A, B and C are throwing a ball to each other. A always throws the ball to B and B throws the ball to C. But C is just as likely to throw the ball to B as to A. If C was the first person to throw the ball, find the probabilities that for the fourth throw (i) A has the ball, (ii) B has the ball, and (iii) C has the ball. Show that, in the long run, A will throw the ball 20 percent of the time, and B and C will throw the ball 40 percent of the time.
Solution
To solve this problem, we can represent the situation using a Markov chain. We'll define the states based on who currently has the ball: , , and .
Transition Probabilities
- From to : 1 (A always throws to B)
- From to : 1 (B always throws to C)
- From :
- To : 0.5
- To : 0.5
Transition Matrix
We can create a transition matrix as follows:
0 & 1 & 0 \\ 0 & 0 & 1 \\ 0.5 & 0.5 & 0 \end{bmatrix}$$ Here, the rows and columns correspond to the states $$[A, B, C]$$. ### State Probabilities Let $$\pi_n = [P(A_n), P(B_n), P(C_n)]$$ be the probabilities of each player having the ball after $$n$$ throws. Initially, since C has the ball: $$\pi_0 = [0, 0, 1]$$ ### Finding Probabilities After 4 Throws We can find the probabilities after each throw by multiplying the current state vector by the transition matrix. 1. **First throw (C to A or B)**: \[ \pi_1 = \pi_0 P = [0, 0, 1] \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0.5 & 0.5 & 0 \end{bmatrix} = [0.5, 0.5, 0] \] 2. **Second throw**: \[ \pi_2 = \pi_1 P = [0.5, 0.5, 0] \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0.5 & 0.5 & 0 \end{bmatrix} = [0, 0.5, 0.5] \] 3. **Third throw**: \[ \pi_3 = \pi_2 P = [0, 0.5, 0.5] \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0.5 & 0.5 & 0 \end{bmatrix} = [0.25, 0.25, 0.5] \] 4. **Fourth throw**: \[ \pi_4 = \pi_3 P = [0.25, 0.25, 0.5] \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0.5 & 0.5 & 0 \end{bmatrix} = [0.25, 0.5, 0.25] \] ### Probabilities After 4 Throws - (i) Probability that A has the ball: $$P(A_4) = 0.25$$ - (ii) Probability that B has the ball: $$P(B_4) = 0.5$$ - (iii) Probability that C has the ball: $$P(C_4) = 0.25$$ ### Long-Term Probabilities To find the long-term (steady-state) probabilities, we solve for $$\pi P = \pi$$: Let $$\pi = [x, y, z]$$. This leads to the equations: 1. $$x = 0.5z$$ 2. $$y = x$$ 3. $$z = y$$ From $$y = x$$ and $$z = y$$, we can write: $$x = 0.5x \Rightarrow 1.5x = 1 \Rightarrow x = \frac{2}{5}, \quad y = \frac{2}{5}, \quad z = \frac{1}{5}$$ So: - $$P(A) = \frac{2}{5}$$ or 40% - $$P(B) = \frac{2}{5}$$ or 40% - $$P(C) = \frac{1}{5}$$ or 20% ### Conclusion In the long run: - A will throw the ball 20% of the time. - B and C will throw the ball 40% of the time each. If you have any questions or need further details, feel free to ask! Here are some follow-up questions for you: 1. Would you like a graphical representation of the transition probabilities? 2. Do you want to explore how the probabilities change with different initial conditions? 3. Would you like to discuss the implications of these probabilities in a real-world scenario? 4. Are there any specific aspects of Markov chains you want to delve into further? 5. Would you like examples of similar problems to practice? **Tip:** Understanding transition matrices is key to solving problems involving Markov chains.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Markov Chains
Transition Matrix
Steady-State Probabilities
Formulas
Transition Matrix P
State Probability Vector π_n = π_0 * P^n
Steady-State Condition π = π * P
Theorems
Markov Chain Steady-State Theorem
Suitable Grade Level
Grades 11-12 or College Level
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