Late For A Date
- Sean and Jess have reservations at Trident at 10am. Sean is walking from Back Bay and Jess is taking the T from Central Square.
- The T is unreliable, so Jess will arrive at some time between 9:50am and 10:10am (whole minutes only) with uniform probability.
- On Sean's way to Trident, there are 10 stores that he likes. He will spend 1 minute in a given store independently with probability \(\frac{1}{3}\). If he doesn't visit any of the stores, he will arrive at Trident at 9:55am (if he goes to all of them, he will be 5 minutes late).
- Their arrival times are independent.
Please answer in fractions of the form x/y.
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Let \(J\) be the random variable for Jess' minute of arrival. I.e. \(J\) gives values in \([-10,10]\). What is the probability that Jess will be exactly on time (i.e. \(\Pr[J=0]\))?
\(J\) has a uniform distribution in \([-10, 10]\), so \(\Pr[J=0]=\frac{1}{10-(-10)+1}=\frac{1}{21}\). -
Let \(S\) be the random variable for Sean's minute of arrival. I.e. \(S\) gives values in \([-5,5]\). What is the probability that Sean will be exactly on time (i.e. \(\Pr[S=0]\))?
We want the probability of Sean visiting exactly 5 stores=5 minutes. Notice that \(T=S+5\) has a binomial distribution with parameters \((10, \frac{1}{3})\). Hence, the answer is given by the general binomial probability function: \(f_{10, \frac{1}{3}}(5)=\binom{10}{5}\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5=\frac{896}{6561}\). -
What is the probability that they will arrive at the same time?
We are asking for the probability that \(J=S\), which is the same as saying that \(J=k\) AND \(S=k\).
\(\Pr[J=S]=\sum_{k=-10}^{10} \Pr[[J=k]\text{ AND }[S=k]]\). Since \(J\) and \(S\) are independent, we get
\(\sum_{k=-10}^{10} \Pr[ J=k \text{ AND } S=k]=\sum_{k=-10}^{10} \Pr[J=k]\Pr[S=k]= \sum_{k=-5}^5 \Pr[J=k]\Pr[S=k]=\\ \frac{1}{21}\sum_{k=-5}^5\Pr[S=k]=\frac{1}{21}\)
We can change the range of the sum because \(\Pr[S=k]=0\) for \(k\notin[-5, 5]\).
\(\sum_{k=-5}^5\Pr[S=k]=1\) since those are all the possible values \(S\) can take.Alternatively, we could have noticed that Sean's spectrum of possible arrival minutes is included in Jess' spectrum, so whatever time Sean arrives, Jess has a \(\frac{1}{20}\) chance of arriving at the same time.