# 4.4 Random Variables, Density Functions

## 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.

1. 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. $$\Pr[J=0]$$)?

Exercise 1

$$J$$ has a uniform distribution in $$[-10, 10]$$, so $$\Pr[J=0]=\frac{1}{10-(-10)+1}=\frac{1}{21}$$.

2. 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. $$\Pr[S=0]$$)?

Exercise 2

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: $$f_{10, \frac{1}{3}}(5)=\binom{10}{5}\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5=\frac{896}{6561}$$.

3. What is the probability that they will arrive at the same time?

Exercise 3

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.