B. Probability Distributions · Discrete Probability DistributionsProbability Histograms and Frequency Histograms
B1.3 B1.1About 65 minutes
Learning goals- I can draw a probability histogram for a discrete random variable.
- I can explain why the total area of the bars in a probability histogram is 1.
- I can compare a frequency histogram from an experiment with the probability histogram from theory.
- I can describe the shape of a distribution (symmetric, skewed, uniform) and identify its mode.
Words to know
probability histogram
A bar graph of a probability distribution: one bar for each value x, centred on x, with height P(X = x). With width 1, each bar's area equals its probability.
frequency histogram
A bar graph of data from an experiment or survey: the height of each bar is how often that value occurred.
relative frequency
Frequency divided by the number of trials. It is an experimental probability.
mode
The most likely value of the random variable: the value with the tallest bar.
skewed
Not symmetric: one tail is longer. Skewed right means the long tail is on the right (toward larger values).
Seeing a distribution
A table of probabilities is exact, but it is hard to see its shape at a glance. A graph shows instantly which values are likely, which are rare, and whether the distribution is balanced.
A probability histogram has:
- the values x of the random variable on the horizontal axis
- one bar for each value, centred on that value, with width 1 (so neighbouring bars touch)
- bar height equal to P(X = x)
Because each bar is 1 unit wide, its area is 1 × P(X = x) = P(X = x). So the area of a bar is the probability of that value, and the total area of all the bars is Σ P(X = x) = 1.
Why area and not just height?
With width-1 bars, height and area are the same number, so it seems like a fussy point. It matters later. In the sibling unit Continuous Distributions and the Normal Distribution, probability is only found as area under a curve. Thinking "probability = area" now makes that step easy.
Frequency histograms come from data
A frequency histogram is drawn from what actually happened in an experiment. Each bar's height is a frequency (a count). If you divide every frequency by the number of trials, you get relative frequencies, which are experimental probabilities. A relative frequency histogram then has the same vertical scale as a probability histogram, and the bars again add to 1. The table after the next graph sums up the differences.
In the unit Probability you saw that experimental probability gets closer to theoretical probability as the number of trials grows. So a relative frequency histogram from many trials should look like the probability histogram. From only a few trials it can look quite bumpy.
| Probability histogram | Frequency histogram |
|---|
| Comes from | theory (a model) | data (an experiment or survey) |
| Bar height | P(X = x) | frequency (or relative frequency) |
| Total area / total of heights | exactly 1 | number of trials (or 1 if relative) |
| Changes if you repeat it? | no | yes, every run is a little different |
Worked exampleExample 1: From data to a relative frequency histogram
The 100 two-dice rolls in the graph gave these frequencies for sums 2 to 12: 1, 7, 11, 11, 17, 20, 11, 11, 5, 4, 2. Find the relative frequency of a sum of 7 and of a sum of 8 or more, and compare with theory.
- Total trials: 1 + 7 + 11 + 11 + 17 + 20 + 11 + 11 + 5 + 4 + 2 = 100. ✓
- Relative frequency of 7: 20/100 = 0.20. Theory: 6/36 ≈ 0.167.
- Sum 8 or more: 11 + 11 + 5 + 4 + 2 = 33, so 33/100 = 0.33.
- Theory: P(X ≥ 8) = (5 + 4 + 3 + 2 + 1)/36 = 15/36 ≈ 0.417.
Experimental: 0.20 and 0.33; theoretical: about 0.167 and about 0.417. The experiment gave more 7s and fewer high sums than theory predicts. That is normal variation in 100 trials, not proof that the dice are unfair.
Worked exampleExample 2: Probability as area
In the two-dice probability histogram, find the total area of the bars from x = 5 to x = 9 and say what it means.
- Each bar has width 1, so area = height.
- Heights: 4/36, 5/36, 6/36, 5/36, 4/36.
- Total area = 24/36 = 2/3.
The area is 2/3, which is P(5 ≤ X ≤ 9): two-thirds of the time, the sum of two dice is between 5 and 9 inclusive.
Describing shape
When you describe a distribution, say:
- Shape: symmetric, skewed right (long tail toward large values), skewed left (long tail toward small values) or uniform (all bars equal).
- Mode: the most likely value (the tallest bar). A distribution with two separate peaks is bimodal.
- Spread: roughly how wide the range of likely values is.
The centre of a distribution, its expected value, is the topic of the next lesson.
Worked exampleExample 3: Drawing and describing a histogram
A bag has 3 red and 2 blue marbles. Draw 2 without replacement; X is the number of red marbles. In lesson 1 you found P(X = 0) = 0.1, P(X = 1) = 0.6, P(X = 2) = 0.3. Describe the probability histogram.
- Draw bars centred at 0, 1 and 2, each of width 1 (from −0.5 to 0.5, 0.5 to 1.5 and 1.5 to 2.5).
- Heights 0.1, 0.6 and 0.3. Areas add to 1. ✓
- The tallest bar is at x = 1, so the mode is 1.
- The bar at 2 is taller than the bar at 0, so the distribution is not symmetric. The longer, lower tail is on the left (toward 0).
Mode 1; slightly skewed left, since red marbles are the majority.
Worked exampleExample 4: Which histogram is which?
Mira rolls one die 30 times and draws a histogram of the results. Her bars have heights 3, 7, 4, 6, 2, 8. Is this a probability histogram? What would the probability histogram look like?
- The heights are counts that add to 30, so this is a frequency histogram.
- For a fair die, P(X = x) = 1/6 for each x = 1 to 6.
- The probability histogram has six equal bars of height 1/6 ≈ 0.167: a uniform shape.
Hers is a frequency histogram from 30 trials. Its bumps are random variation. The probability histogram is flat (uniform) at height 1/6.
Just for funRoll two dice 1000 times and someone will always say the 7s are "due" after a few rolls without one. Dice have no memory, and they never read the probability histogram either.
Hands-on activityYour own dice experiment
You need: two dice (or a free online dice roller, or two sets of slips of paper numbered 1 to 6 in two cups), pencil, paper and graph paper or a spreadsheet.
- Roll the two dice 60 times and tally each sum from 2 to 12.
- Draw a frequency histogram of your results, with bars of width 1 that touch.
- Divide each frequency by 60 to get relative frequencies. Draw a relative frequency histogram on the same scale as the probability histogram in this lesson.
- Describe the shape and the mode of your graph. Compare it with the probability histogram.
- Combine your results with a friend's or family member's (or roll 60 more times). Does the combined graph look closer to the theory?
What to look for: the experimental mode might not be 7, and some bars may be far from the theory with only 60 rolls. The combined 120 rolls should usually look more symmetric and closer to the peaked shape of the theoretical histogram. In a spreadsheet, =RANDBETWEEN(1,6)+RANDBETWEEN(1,6) simulates one roll of two dice.
Your turn
1A probability histogram has bars at x = 0, 1, 2, 3 with heights 0.1, 0.3, 0.4 and h. Find h.
The areas must add to 1: 0.1 + 0.3 + 0.4 + h = 1, so h = 0.2.
2For the histogram in the previous question, what is the mode, and what is P(X ≥ 2)?
Mode 2 (tallest bar, 0.4). P(X ≥ 2) = 0.4 + 0.2 = 0.6.
3Toss 3 coins; X = number of heads. Describe the probability histogram.
Bars at 0, 1, 2, 3 with heights 1/8, 3/8, 3/8, 1/8. Symmetric, with two equal tallest bars at 1 and 2.
4In 200 rolls of one die, a 4 came up 41 times. What is the relative frequency of 4, and how does it compare with the theoretical probability?
41/200 = 0.205; theory is 1/6 ≈ 0.167. It is higher, but a difference like this is not unusual in 200 rolls.
5Why is the total area of a probability histogram always 1?
Each bar has width 1, so its area equals P(X = x). The total area is the sum of all the probabilities, which is 1 for any distribution.
6A frequency histogram from 50 trials has bar heights 5, 10, 20, 15. Convert to relative frequencies.
Divide by 50: 0.10, 0.20, 0.40, 0.30.
7A distribution has probabilities 0.40, 0.30, 0.15, 0.10, 0.05 for x = 0 to 4. Describe its shape.
The tallest bar is at 0 and the bars shrink to the right: skewed right, mode 0.
More practice: warm-up
1In a probability histogram, what does the height of a bar show?
P(X = x), the probability of that value.
2In a frequency histogram, what does the height of a bar show?
The frequency: how many times the value happened.
3What is the total area of a probability histogram?
4What is the mode of the sum of two dice?
5Four equal bars of height 0.25. What shape is this?
More practice: on your own
1Use the two-dice histogram. Find the area of the bars for x ≤ 4 and for x ≥ 10, and explain why they are equal.
x ≤ 4: (1 + 2 + 3)/36 = 6/36. x ≥ 10: (3 + 2 + 1)/36 = 6/36. They are equal because the histogram is symmetric about 7.
2Roll two dice and let D be the absolute difference. The probabilities are 6/36, 10/36, 8/36, 6/36, 4/36, 2/36 for d = 0 to 5. Describe the shape and the mode.
Mode 1 (10/36). After the peak the bars fall steadily to the right: skewed right.
3A student runs 500 trials of an experiment and the relative frequency histogram looks very different from the theoretical one. Give two possible explanations.
Possible answers: the model is wrong (for example the dice are unfair or the outcomes are not equally likely); there was an error in recording or in the theory; or, less likely with 500 trials, an unusual run of chance.
4A probability histogram has heights c, 2c, 3c, 2c, c for x = 1 to 5. Find c and P(2 ≤ X ≤ 4).
9c = 1, so c = 1/9. P(2 ≤ X ≤ 4) = (2 + 3 + 2)/9 = 7/9.
5Why should you not draw a probability histogram with gaps between the bars, if you want area to equal probability?
With width 1 and no gaps, each bar spans from x − 0.5 to x + 0.5, so area = height = P(X = x). Narrower bars with gaps would have areas smaller than the probabilities, and the total area would be less than 1.
Challenge
1Spin a spinner numbered 1, 2, 3, 4 (equal sectors) twice and let M be the larger of the two numbers. Find the distribution, describe the histogram and find the area to the right of x = 2.5.
16 outcomes. M = k in k² − (k − 1)² = 2k − 1 ways: 1, 3, 5, 7. P = 1/16, 3/16, 5/16, 7/16. Bars rise to the right: skewed left, mode 4. Area right of 2.5 = P(M ≥ 3) = 12/16 = 3/4.
2In 360 rolls of two dice, about how many 7s and how many 12s would you expect? Is it surprising to get 55 sevens?
Expected about 360 × 6/36 = 60 sevens and 360 × 1/36 = 10 twelves. 55 sevens is only 5 fewer than expected, which is not surprising for random variation.
3Two histograms are drawn for X = number of heads in 10 coin tosses: one from theory, one from 20 runs of the experiment. Explain why the experimental one might have no bar at all at x = 0 or x = 10, and whether the theoretical one has bars there.
Theory: P(X = 0) = P(X = 10) = (1/2)¹⁰ = 1/1024, so the theoretical bars exist but are tiny. In only 20 runs, an event with probability about 0.001 will almost surely not happen, so the experimental histogram usually has no bar at the ends.
Quiz
1. In a probability histogram with bars of width 1, the area of a bar equals...
2. A relative frequency histogram from 1000 trials will most likely...
3. Bars at x = 1, 2, 3, 4 have heights 0.5, 0.25, 0.15, 0.10. How is the distribution best described?
4. A spinner experiment gives frequencies 12, 18, 30 for x = 1, 2, 3. What is the relative frequency of x = 3?
5. What is the mode of the distribution P(X = 0) = 0.2, P(X = 1) = 0.35, P(X = 2) = 0.45?
6. Which statement about a frequency histogram is true?
Big idea (B1.3): a probability histogram is a picture of a probability distribution. Each bar is centred on a value, has width 1, and its area is that value's probability, so the bars add to 1. A frequency histogram is drawn from real data and only approximates the theoretical shape.
How to help: roll dice together and build a tally chart, then compare your graph with the one in the lesson. Ask: "Which graph is the theory and which is what actually happened? Would they match better with more rolls?"
Common difficulties:
- Leaving gaps between bars or placing bars between values instead of centred on them.
- Giving the probability (0.45) when asked for the mode (the value 2).
- Reading skew backwards: "skewed right" means the tail is on the right, not the peak.
- Expecting a small experiment to match theory exactly.