Subjects probability

Bus Late Probability

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Bus Late Probability


1. **Problem Statement:** We are given probabilities of weather conditions and the probability of a bus being late under each condition. We need to: (a) Represent this information in a tree diagram. (b) Calculate the probability that the weather is dry and the bus is on time. (c) Calculate the overall probability that the bus is late. 2. **Given Data:** - Probability of dry weather, $P(D) = \frac{3}{4}$ - Probability of rainy weather, $P(R) = \frac{1}{5}$ - Probability of snowy weather, $P(S) = \frac{1}{20}$ - Probability bus is late given dry weather, $P(L|D) = \frac{1}{10}$ - Probability bus is late given rainy weather, $P(L|R) = \frac{1}{4}$ - Probability bus is late given snowy weather, $P(L|S) = \frac{2}{3}$ 3. **Tree Diagram Explanation:** The tree starts with three branches for weather: Dry, Rainy, Snowy with probabilities $\frac{3}{4}$, $\frac{1}{5}$, and $\frac{1}{20}$ respectively. Each weather branch splits into two branches: Bus Late (L) and Bus On Time (T). - For Dry: $P(L|D) = \frac{1}{10}$, so $P(T|D) = 1 - \frac{1}{10} = \frac{9}{10}$ - For Rainy: $P(L|R) = \frac{1}{4}$, so $P(T|R) = \frac{3}{4}$ - For Snowy: $P(L|S) = \frac{2}{3}$, so $P(T|S) = \frac{1}{3}$ 4. **Calculations:** (a) Tree diagram is conceptual as described above. (b) Probability weather is dry and bus is on time: $$ P(D \cap T) = P(D) \times P(T|D) = \frac{3}{4} \times \frac{9}{10} = \frac{27}{40} = 0.675 $$ (c) Probability bus is late (total probability): $$ P(L) = P(D)P(L|D) + P(R)P(L|R) + P(S)P(L|S) = \frac{3}{4} \times \frac{1}{10} + \frac{1}{5} \times \frac{1}{4} + \frac{1}{20} \times \frac{2}{3} = \frac{3}{40} + \frac{1}{20} + \frac{1}{30} $$ Find common denominator 120: $$ \frac{3}{40} = \frac{9}{120}, \quad \frac{1}{20} = \frac{6}{120}, \quad \frac{1}{30} = \frac{4}{120} $$ Sum: $$ P(L) = \frac{9}{120} + \frac{6}{120} + \frac{4}{120} = \frac{19}{120} \approx 0.1583 $$ **Final answers:** - (b) $P(D \cap T) = \frac{27}{40} = 0.675$ - (c) $P(L) = \frac{19}{120} \approx 0.1583$ This completes the solution with the tree diagram structure and probability calculations.