The full quiz is here. The answers appear below. Comments, which are not part of the answers, are italicized.
"A bag contains a bean that is known either to be white or black. A white bean is added to the bag, the bag is shaken, and one bean is taken out at random. It is white. What, now, is the chance that the bean left in the bag is white?"
Suppose you do some further research and ascertain that, at the beginning of the problem (before adding the white bean), the probability that the bag contains a black bean is 2/3.
1. This problem concerns a sequence of events: place a bean into a bag; after adding the white bean, draw one bean, then observe the remaining bean. Write down all possible sequences and determine their probabilities.
We may suppose that a bean is originally drawn from a box containing two black beans and one white bean. That's how we model probabilities like 2/3. You could just as well assume there are, say four black beans and two white beans, or any other number in a 2:1 proportion, but that would just make you work harder. This bean is put into the bag. A white bean is added, the bag is shaken, and a white bean is taken out. We observe the color of the bean left in the bag.
Let us label the two black beans in the box: we will call them B1 and B2. We will call the white bean in the box W1. The white bean added to the bag is W. Here, then, are all the possible sequences, including those that do not occur:
| Number | Bean drawn from box | Beans put into bag | Bean drawn from bag | Bean left in bag |
| 1 | W1 | W1, W | W1 | W |
| 2 | W1 | W1, W | W | W1 |
| 3 | B1 | B1, W | B1 (does not occur) | W |
| 4 | B1 | B1, W | W | B1 |
| 5 | B2 | B2, W | B2 (does not occur) | W |
| 6 | B2 | B2, W | W | B2 |
Each row has an equal probability of occurring. Beans W1, B1, and B2 are drawn from the box with equal probability. One of the two beans is drawn from the bag with equal probability. The draws are independent. Therefore the six combinations of two draws are equiprobable.
2. Model the sequences of outcomes using a single box model.
The box contains six tickets, one for each row of the table. Each ticket shows the color of the bean drawn from the bag and the color of the bean left in the bag. These are the rightmost two columns of the table (with the suffixes "1" and "2" removed). Therefore the tickets are
WW, WW, BW, WB, BW, and WB.
It is acceptable at this point to remove the two BW tickets from the box, because we are told they will not be drawn. Therefore another correct answer is that the box contains just four tickets, WW, WW, WB, and WB. Because only the proportions matter, it is also correct to answer that the box contains just two tickets, WW and WB. However, it is important that you show how you derived the box contents: it would not be correct simply to state the answer without any kind of analysis.
3. Answer the question: "What, now, is the chance that the bean left in the bag is white?"
When we are told that the bean taken out of the bag is white, in terms of the box model this says we know that the ticket drawn is either #1, #2, #4, or #6: that is, it is one of WW, WW, WB, and WB, each with equal probability. Of these tickets, two show a white bean left in the bag. Therefore the probability is 2/4 = 50%.
4. Write down an Excel expression that will be equal to -1 10% of the time (at random) and equal to 1 90% of the time.
There are many ways to do this. One of the most straightforward is IF(RAND() < 0.10, -1, 1). Another approach is to use arithmetic, as in INT(10*RAND()/9)*(-2) + 1. 10*RAND()*9 will be between 0 and 1 9/10 = 90% of the time and between 1 and 10/9 the remaining 10% of the time. The INT() function converts the values to 0 and 1; multiplying by -2 and adding 1 converts those to 1 and -1, respectively. Therefore this expression is equal to 1 90% of the time and to -1 the remaining 10% of the time.
Scoring: The passing score is 90.
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This page was created 10 February 2001.