Question

Probability of trains A, B and C arriving on times is 1/3, 2/5 and 2/3 respectively. What is the probability that almost one of the trains arrive on time?

a) 7/15 b) 22/45 c) 23/45 d) None of these

Answer 1

$$\begin{gathered} \mathrm{Dear}\mathrm{Student}, \\ \mathrm{Probability}\mathrm{of}\mathrm{train}\mathrm{A}\mathrm{arriving}\mathrm{on}\mathrm{time}=\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{3} \\ \mathrm{Thus},\mathrm{Probability}\mathrm{of}\mathrm{train}\mathrm{A}\mathrm{not}\mathrm{arriving}\mathrm{on}\mathrm{time}=\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)=1-\frac{1}{3}=\frac{2}{3} \\ \mathrm{Similarly}, \\ \mathrm{P}\left(\mathrm{B}\right)=\frac{2}{5}\mathrm{and}\mathrm{P}\left({\mathrm{B}}^{\mathrm{c}}\right)=\frac{3}{5} \\ \mathrm{P}\left(\mathrm{C}\right)=\frac{2}{3}\mathrm{and}\mathrm{P}\left({\mathrm{C}}^{\mathrm{c}}\right)=\frac{1}{3} \\ \mathrm{Probability}\mathrm{that}\mathrm{ATMOST}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{trains}\mathrm{arrive}\mathrm{on}\mathrm{time}= \\ \mathrm{Probability}\mathrm{that}\mathrm{none}\mathrm{of}\mathrm{them}\mathrm{arrive}\mathrm{on}\mathrm{time}+\mathrm{Probability}\mathrm{that}\mathrm{only}\mathrm{train}\mathrm{A}\mathrm{arrive}\mathrm{on}\mathrm{time}+ \\ \mathrm{Probability}\mathrm{that}\mathrm{only}\mathrm{train}\mathrm{B}\mathrm{arrive}\mathrm{on}\mathrm{time}+\mathrm{Probability}\mathrm{that}\mathrm{only}\mathrm{train}\mathrm{C}\mathrm{arrive}\mathrm{on}\mathrm{time} \\ =\left[\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)\times \mathrm{P}\left({\mathrm{B}}^{\mathrm{c}}\right)\times \mathrm{P}\left({\mathrm{C}}^{\mathrm{c}}\right)\right]+\left[\mathrm{P}\left(\mathrm{A}\right)\times \mathrm{P}\left({\mathrm{B}}^{\mathrm{c}}\right)\times \mathrm{P}\left({\mathrm{C}}^{\mathrm{c}}\right)\right]+\left[\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)\times \mathrm{P}\left(\mathrm{B}\right)\times \mathrm{P}\left({\mathrm{C}}^{\mathrm{c}}\right)\right]+\left[\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)\times \mathrm{P}\left({\mathrm{B}}^{\mathrm{c}}\right)\times \mathrm{P}\left(\mathrm{C}\right)\right] \\ =\left[\frac{2}{3}\times \frac{3}{5}\times \frac{1}{3}\right]+\left[\frac{1}{3}\times \frac{3}{5}\times \frac{1}{3}\right]+\left[\frac{2}{3}\times \frac{2}{5}\times \frac{1}{3}\right]+\left[\frac{2}{3}\times \frac{3}{5}\times \frac{2}{3}\right] \\ =\frac{6}{45}+\frac{3}{45}+\frac{4}{45}+\frac{12}{45} \\ =\frac{25}{45} \\ =\frac{5}{9} \\ \mathrm{Regards} \end{gathered}$$