Demonstration 1 of 4
Zero and one are the walls
On one fair die, how small and how large can the probability of an event get?
The probability is the favorable count over 6. That count can never be below 0 or above 6, so the probability stays between 0 and 1. The event and its complement split the six faces, so their probabilities always add to 1.
\[0\le P(E)\le1\]
\[2/6=1/3\]
E is an event: a stated collection of outcomes. P(E) means the probability of event E. The six faces 1 through 6 are the equally likely outcomes. The complement of an event holds every face that is not in it.
Predict first. Change the event to roll more than 6. How many faces belong now, and what is the probability?
Choose an example
Constructed example: the chapter's fair die and its events more than 4, 8, and less than 7.
Calculated values
- Favorable faces
- 5, 6
- P(more than 4)
- 2/6 = 1/3
- P(at most 4)
- 4/6 = 2/3
- Event + complement
- 1/3 + 2/3 = 1
Favorable faces: 5, 6. P(more than 4) = 2/6 = 1/3. The complement, at most 4, has 4/6 = 2/3. Check: 2/6 + 4/6 = 6/6 = 1, because every face is in exactly one of the two events. The probability sits strictly between 0 and 1.
Use the idea
When a probability claim comes out negative or above 1, stop and recount. Something was left out, counted twice, or divided by the wrong total.
Where the conclusion applies
One ideal fair die numbered 1 through 6, so each face has the same chance. The counting shortcut needs that equal chance: a spinner with three quarters of its area blue would not give two colors equal probability.
Check your understanding: On the same die, what is the probability of rolling at most 2, and of its complement?
Chapter 21 source: section "Zero and one mark the probability boundaries". Demonstration C21-D01.
Demonstration 2 of 4
Eleven sums, thirty-six pairs
Two fair dice can show sums from 2 to 12. Why is 7 more likely than 2?
Every cell in the grid is one ordered pair with chance 1/36. Choosing a sum outlines the cells that add to it. Middle sums have more cells, so they collect more probability, even though every sum is one label.
\[6 × 6 = 36\]
\[6/36=1/6\]
An ordered pair (die 1, die 2) records which die showed which value, so (2, 5) and (5, 2) are different outcomes. The 36 ordered pairs are the equally likely outcomes. A sum is an event: the collection of pairs that add to it.
Predict first. Before you switch to the sum 12, guess how many cells will be outlined.
Choose an example
Constructed example: the chapter's two ideal labeled dice and its sums 2, 7 and 12.
Calculated values
- Ordered pairs in the grid
- 6 x 6 = 36
- Pairs with sum 7
- 6: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
- P(sum 7)
- 6/36 = 1/6
- Possible sums
- 11 labels, 2 through 12
Each cell shows its sum. The sum 7 collects 6 of the 36 equally likely ordered pairs: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). So P(sum 7) = 6/36 = 1/6. Each pair has (1/6) x (1/6) = 1/36, and 6 x 1/36 = 6/36. There are 11 sum labels, but they are not 11 equal shares.
Use the idea
Before dividing by a count of possible results, ask whether those results are equally likely. If they are not, go down to outcomes that are, and count those.
Where the conclusion applies
Two ideal labeled dice that are fair and independent, which makes all 36 ordered pairs equally likely. If the two dice were coupled, the pairs would not all have chance 1/36 and the count would fail.
Check your understanding: How many ordered pairs give a sum of at least 10, and what is its probability?
Chapter 21 source: section "Equal elementary outcomes can produce unequal events". Demonstration C21-D02.
Demonstration 3 of 4
Take a token out of the bag
Does the first draw change the chance of red on the second draw?
The second draw uses whatever is left in the bag. Taking a token out changes the bag, so the second chance depends on the first color. Putting it back restores the original bag, so the chance stays 2/3.
\[2/3\]
\[1/2\]
The bag holds two red tokens (R) and one blue token (B). P(red) is the chance the next draw is red. Without replacement, the first token stays out; with replacement, it goes back and the bag is mixed again.
Predict first. Keep the token out and switch the first draw to blue. What is the chance of red on the second draw?
Choose an example
Constructed example: the chapter's bag of two red tokens and one blue token.
Calculated values
- P(red on draw 1)
- 2/3
- Tokens for draw 2
- 1 red, 1 blue
- P(red on draw 2 after red)
- 1/2
- P(red on draw 2 after blue)
- 1
- Does draw 2 depend on draw 1?
- yes
Before any draw, P(red) = 2/3. The red token stays out, leaving 1 red and 1 blue: P(red) = 1/2. After blue it would be 1, so the draws are dependent.
Use the idea
For any repeated draw, write down what remains after the first result before you compute the second chance.
Where the conclusion applies
Each token left in the bag is equally likely to be drawn. The model fails if, say, the tokens are not mixed after one is returned.
Check your understanding: A bag holds three green and two yellow tokens. A green token is drawn and kept out. What is the chance the next token is green?
Chapter 21 source: section "Worked example: removing an object changes the next trial". Demonstration C21-D03.
Demonstration 4 of 4
Eight paths of three flips
In three fair flips, how likely is exactly two heads?
The tree lists all eight ordered paths. Choosing a number of heads highlights the paths with that many H letters. HHT and THH have the same counts but are different paths, and each one counts.
\[2 × 2 × 2 = 8\]
\[(1/2)(1/2)(1/2) = 1/8\]
H is heads and T is tails. A path such as HTH lists the three flips in order. Each branch has probability 1/2, so each complete path has (1/2)(1/2)(1/2) = 1/8.
Predict first. Switch to exactly one head. Will it have more paths, fewer, or the same number as two heads?
Choose an example
Constructed example: the chapter's three-flip sample space and Figure 21.2.
Calculated values
- Paths in the sample space
- 2 x 2 x 2 = 8
- Probability of each path
- (1/2)(1/2)(1/2) = 1/8
- Paths with exactly two heads
- HHT, HTH, THH
- P(exactly two heads)
- 3/8
- Head counts 0, 1, 2, 3
- 1 + 3 + 3 + 1 = 8
The event exactly two heads keeps the leaves HHT, HTH, THH. That is 3 of 8 equally likely paths, so P(exactly two heads) = 3 x 1/8 = 3/8. Each path has (1/2)(1/2)(1/2) = 1/8 because the flips are fair and independent. Check the whole tree: 1 + 3 + 3 + 1 = 8.
Use the idea
Check an event two ways: once by listing outcomes and once by following a tree or grid. If the two counts disagree, something was skipped or counted twice.
Where the conclusion applies
Three ideal fair coin flips that are independent, so every path has chance 1/8. A bent coin, or flips that influence each other, would make the paths unequal.
Check your understanding: In two fair independent flips, what is the probability of exactly one head?
Chapter 21 source: section "Build the three-flip sample space". Demonstration C21-D04.