EX 2.15 select car of train

• During off-peak hours, a commuter train has five cars.
• Suppose a commuter
  • is twice as likely to select the middle car (#3) as to select either adjacent car (#2 or #4),
  • is twice as likely to select either adjacent car as to select either end car (#1 or #5).
• Let $p_i$ = $P($ car $i$ is selected $)$ = $P\left(E_i\right)$.
• Then we have
  • $p_3=2p_2=2p_4$
  • $p_2=2p_1=2p_5=p_4$.
• This gives

$$1=\sum P\left(E_i\right)=p_1+2p_1+4p_1+2p_1+p_1=10p_1$$

• It implies
  • $p_1=p_5=.1$,
  • $p_2=p_4=.2$,
  • $p_3=.4$.
• The probability that one of the three middle cars is selected (a compound event) is then $p_2+p_3+p_4=.8.$

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