Assignment 2 - Conditional Probability

Consider the experiment that consists of tossing four coins (or, equivalently, of tossing a single coin four times).

This experiment has sixteen equally likely outcomes:

In the first assignment, we estimated the probability of an event using the empirical approach, which means we repeat the experiment and see what proportion of the time the event actually occurs.

In this exercise we will use the empirical approach to estimate a conditional probability of an event. A conditional probability is just the probability of an event occurring on the condition that we know some other event occurred.

Example: If the experiment consists of tossing a fair coin four times, there are 16 outcomes in the sample space, and four of them result in the event "three heads". We assume the sixteen outcomes are equally likely, so each has probability 1/16 and "three heads" has probability 4/16 or 1/4.

Now suppose we are given the additional information that the first toss came up tails. Knowing this alters the probability of three heads, because now we are restricted to only the last 8 outcomes in the table, only one of which produces three heads. Since there are now eight outcomes, only one of which produces three heads, we suspect (correctly) that the conditional probability of three heads when we know that the first toss is tails is 1/8.

The formula for conditional probability given on page 257 in the text allows us to determine the conditional probability of an event in a general way. We will use the empirical method to estimate some conditional probabilities and compare our estimates to the result of the the formula.

Directions for the Assignment

We want to estimate the following conditional probabilities for the four coin experiment:

1. What is the conditional probability of getting exactly two heads, given that the first toss came up heads?
2. What is the conditional probability that the first toss came up heads, given that exactly two heads came up in four tosses?

• Make a spreadsheet that has four columns, each one with a simulated coin toss (You can modify the Twocoins spreadsheet, or your Assignment 1 spreadsheet for this). Make the number of rows fairly large, say several hundred.
• Add a column (say, E) that contains one if exactly two heads came up, zero otherwise. You can use the IF function together with the SUM function to accomplish this.
• Add a column (F) that contains one if toss 1 (Column A) came up 1 (heads), and exactly two heads came up (i.e., column E contains one). You can use the AND function in conjunction with IF for this.

Question 1

We only want to consider outcomes where column A contains a 1, and from this group we want to count the number of times four tosses produced exactly two heads. Column F contains the number of times both events, "heads on toss 1" and "exactly two heads" occurred. We can estimate the probability of both events occurring from the proportion of ones in column F.

However, we are not quite done. According to the conditional probability formula, we have to divide this result by the probability of the event we are conditioning on, "heads on toss 1". We can estimate this probability from the proportion of ones in column A.

So, our estimated conditional probability is:

(Proportion of ones in Column F) divided by (Proportion of ones in Column A)

The proportions can be computed by using the SUM function to add up each of the columns A and F, and dividing by the number of rows. Once we have the proportions (say, in a cell in the first row of that column), we can make a formula like =F1/A1 to compute the ratio of the two proportions. This ratio is our estimate of the conditional probability of exactly two heads given that the first toss came up heads.

Question 2

This time we only want to consider outcomes where column E contains a 1, meaning exactly two heads came up. and from this group we want to count the number of times the first toss was heads (Column A contains 1). As before, we can estimate the probability of both events occurring from the proportion of ones in column F.

This time we are conditioning on the event "exactly two heads", so we have to divide the probability of both events (two heads and heads on toss 1) occurring by the probability of the event we are conditioning on, "exactly two heads". We can estimate this probability from the proportion of ones in column E.

So, our estimated conditional probability is:

(Proportion of ones in Column F) divided by (Proportion of ones in Column E)

Discussion

1. Exactly two heads come up in four tosses
2. The first toss comes up heads
3. Both 1) and 2) occur simultaneously

To determine the probabilities of each of these events, use the classical method: count the number of rows in the table of outcomes that correspond to the event, and divide by the number of outcomes (16).