Question

Say I'm on a date. The probability of my date showing up on time for the first date is 0.5 (since possible outcomes are either “on time” or “not on time”, the probability of being “not on time” is 1 – P (on time) = 1 - 0.5 = 0.5). For the second date, the probability of the same person showing up on time is 0.90 if she was on time for the first date. However, this probability is only 0.35 if she was not on time for the first date. Show the work.

1. Show the probability tree summarizing probabilities of the potential outcomes of these two dates, showing up on time or not.

2. Suppose punctuality is important to me, and if my date is late both times, so I have no interest in a relationship. What is the probability of that happening?

3. Use the probability tree from #1 to construct a discrete probability distribution for the potential outcomes in terms of number of times my date showed up on time over the two dates.

4. From the discrete probability distribution constructed in #3 calculate the mean and variance of the potential outcomes, i.e. the number of times my date showed up on time.

Answer 1

Suppose, denote the events that the date shows up on time and does not show up on time for the first and second dates respectively. Then,

Since there are only two possible outcomes of showing up on time or not showing up on time in each date, we have,

Now,

Using these probabilities we construct the probability tree:

The probability that the date is late both times,

If X denotes a random variable denoting the number of times the date showed up on tim over the two dates, then he Probability Distribution of X is given by the following:

X=x	Events	Pr[X=x]
0		0.325
1		0.050+0.175=0.225
2		0.450
	TOTAL	1

Next,