Fall Semester 2002

Lab Five, Wednesday November 13, 2002
Date
Nov 13, 2002

Due today
Tutorial Four
Last day to turn in Homework Two with no penalty.
Please check that you can get into QuizSite.

Starting today
Excel Tutorial Five (from the lab manual).
There's a problem and an experiment listed below too.

Due next time
Excel Tutorial Five
Prepare an experiment like the one listed below.

Today's grading
You might receive questions of the following nature:
  1. How do you create an embedded chart?
  2. How do you edit an embedded chart?
  3. How do you select non-adjacent ranges?
  4. If you throw two dice, what's the probability of getting two fours?
  5. If you throw two dice, what's the probability of getting at least one four?
  6. If you throw two dice, what's the probability of getting exactly one four?
  7. If you throw two dice, what's the probability of getting anything but fours?

Today's custom experiment
The Two Gold Coins Problem

There is a box with three drawers, one with two gold coins, one with a gold and a silver coin, and one with two silver coins. A drawer is chosen at random and then a coin in the drawer is chosen at random. The observed coin is gold. What is the probability that the other coin is also gold?

Here's a solution: The original sample space before the observation is clearly the product space of the three random drawers and the two random choices of which coin (in order to count carefully we will give the coins marksa 1 and 2 when there are two of the same kind as illustrated below).

Order
first/second first/second
drawer 1 G1G2 G2G1
drawer 2 G S S G
drawer 3 S1S2 S2S1
These six cases, choice of drawer (p = 1/3), and then choice of which coin (p = 1/2), exhaust the sample space. Thus, each compound event has a probability of 1/6. (The events are independent).

In this sample space the probability of drawing a gold coin on the first draw (or on the second) is 1/2. However, the observation that the first drawn coin is gold eliminates the three blue possibilities.

The remaining thre points in the sample space:

are still equally likely since originally each had the same probability 1/6 so now each has the conditional probability of
(1/6) / (1/6 + 1/6 + 1/6) = 1/3
which results from dividing the successes by the total in the reduced sample space.

Of these three possibilities two give a gold coin on the second draw and only one gives a silver coin, hence the probability of a second gold coin is 2/3. End of problem.

Conclusion

Frequently the probability we want to know is conditional on some event. The effect of the condition is to remove some of the events in the sample space (list), or equivalently to confine the admissible items of the sample space to a limited region. (That's what we did towards the end of the discussion, above).

Second, you don't have to either like or to believe (or even accept the above argument.) If this result seems strange to you (after all the second coin is either G or S so why not p = 1/2?), then think through how you would do a corresponding experiment.

Your task, therefore, is to confirm or refute (through an experiment) the probability calculated above. Carefully design, then carry out the experiment. What is the probability value you come to?

Last updated: Oct 31, 2002 by Adrian German for A113