Random variables and probability distributions, reading a most likely value and a range from a histogram, the conditions of a binomial experiment and the formula for exactly k successes, constructing and interpreting a binomial distribution, and describing a distribution as symmetric or skewed.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 10 — Counting Methods and Probability
Construct and Interpret Binomial Distributions
Objectives
Five outcomes. Every possible result, with its probability, in one picture.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 724-731 — the lesson these objectives are drawn from
Warm-up
Lesson 10.3 found the probability of a single event; Lesson 10.5 handled repeated independent trials.
Discussion prompt
Rolling two dice, you can find the probability of a sum of 7. What would you learn by finding the probability of every possible sum at once?
Hint: Think about what a full list would show.
Answer:
You would see which sums are likely and which are rare, all in one place — and you could answer any question about sums by adding the right entries.
\[ P(2)+P(3)+\dots+P(12) = 1 \]
That complete list is a probability distribution, and the picture of it is a histogram. This lesson builds them and reads them.
Concept
A random variable takes a value determined by a random event. A probability distribution gives the probability of each possible value, and those probabilities must total 1.
binomial distribution — The probability distribution for the number of successes in n independent trials, each with two outcomes and the same success probability p.
\[ P(k) = \;_nC_k\,p^k(1-p)^{n-k} \]
When the situation is repeated independent trials with two outcomes and a fixed success rate, every probability in the distribution comes from a single formula.
Figure (svg): Two columns comparing a general probability distribution with a binomial one
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 724-725
Section
Section 1
Concept
A random variable's probability distribution lists each possible value with its probability. Displayed as a histogram, the height of each bar is that value's probability, and all the heights together account for 1.
\[ \sum P(X) = 1 \]
The requirement that the probabilities total 1 is a free check on every distribution: if they do not, a value has been missed or a probability miscomputed.
Figure (svg): A histogram of the probability distribution for the sum of two dice
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 724-724 — Probability Distributions
Picture it
Example 1: the sum of two dice.
Figure (svg): A histogram of the probability distribution for the sum of two dice
The bars rise to a peak at 7 and fall away symmetrically, because there are more ways to make a middling sum than an extreme one.
Worked example
Example 1.
\[ \text{Let } X \text{ be the sum of two six-sided dice. Make its probability distribution.} \]
List the possible values
Why: Every sum from 2 to 12.
\[ 11\text{ values} \]
Count the outcomes for each
Why: One way to make 2, two ways to make 3, and so on.
\[ 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 \]
Divide each by 36
Why: There are 36 equally likely pairs.
\[ P(7) = \frac{6}{36} = \frac{1}{6} \]
Check the total
Why: The eleven counts add to 36.
\[ \text{the probabilities total } 1 \]
Figure (svg): A histogram of the probability distribution for the sum of two dice
\[ P(X) = \tfrac{\text{outcomes}}{36} \]
Verify: check the counts total 36
Why: One plus 2 plus 3 plus 4 plus 5 plus 6 plus 5 plus 4 plus 3 plus 2 plus 1 is 36, matching the number of dice pairs. Since every pair produces exactly one sum, the counts must account for all 36 — and that is what makes the probabilities total 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 724-724
Fill the middle
Example 1.
Fill in the blanks
P(X = 7) = \frac6}___ = \frac______
Why: Six of the 36 pairs sum to 7, which is why it is the tallest bar. Every other sum has fewer pairs producing it.
Worked example
Guided Practice 1.
\[ \text{Repeat for two tetrahedral dice numbered } 1 \text{ to } 4. \]
List the possible values
Why: Sums from 2 to 8.
\[ 7\text{ values} \]
Count the outcomes
Why: One, 2, 3, 4, 3, 2, 1 ways.
\[ 16\text{ pairs in total} \]
Divide each by 16
Why: Four times 4 equally likely pairs.
\[ P(5) = \frac{4}{16} = \frac{1}{4} \]
Check the total
Why: The seven counts add to 16.
\[ \text{probabilities total } 1 \]
Figure (svg): The solution to Worked example four-sided dice shown as a ladder of expressions, one row per algebraic move
\[ P(5) = \tfrac{4}{16} = \tfrac{1}{4} \]
Verify: compare the shape with the six-sided case
Why: Both rise to a single peak in the middle and fall away symmetrically — 4 out of 16 at the top here, 6 out of 36 there. The shape comes from the counting rather than from the number of faces, so any pair of fair dice produces the same triangular profile.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 725-725
Trap
\[ 11 \text{ possible sums, from } 2 \text{ to } 12 \]
Assign each sum a probability of one eleventh
Why: The values are listed and treated as equally likely.
\[ P(7) = \tfrac{1}{11} \quad \text{(wrong)} \]
Six of the 36 pairs give a sum of 7 but only one gives a sum of 2, so the sums are very far from equally likely.
\[ P(7) = \tfrac{6}{36} = \tfrac{1}{6}; \quad P(2) = \tfrac{1}{36} \]
Count the outcomes producing each value
Why: The equally likely items are the 36 pairs, not the 11 sums.
\[ \text{the histogram is a triangle, not a flat line} \]
Identifying what is genuinely equally likely is the first step of every distribution, and it is where this whole lesson's arithmetic rests.
Sorting
Count the pairs producing it.
Sort into buckets
Sort each sum of two dice.
The number of ways rises to a peak at 7 and falls away symmetrically, which is exactly the triangular shape of the histogram.
Matching
Out of 36 pairs.
Match the pairs
Why: A sum of 10 has the same probability as a sum of 4, and 5 the same as 9 — the distribution is symmetric about 7, since each sum s pairs with 14 minus s.
Prediction
Commit before reasoning.
Predict first
Why do the probabilities in any distribution add to exactly 1?
Correct: Because the values are exhaustive and mutually exclusive, so something must happen.
\[ 1+2+3+4+5+6+5+4+3+2+1 = 36 \]
Why: Every outcome produces exactly one value of the variable, so the counts partition the whole sample space and the probabilities partition 1. It holds for every distribution, however the probabilities were obtained. That makes the total a free check: if a computed distribution does not sum to 1, something is missing or wrong.
Section
Section 2
Concept
The most likely value is the one with the tallest bar. The probability that the variable falls in a range is the sum of the probabilities of the values in it, since different values cannot happen together.
\[ P(X \geq 10) = P(10)+P(11)+P(12) \]
Adding is legitimate because the values are disjoint by construction: the variable takes exactly one of them.
Figure (svg): The dice histogram with the tallest bar and a tail region marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 725-725 — Interpret a probability distribution
Picture it
Example 2: the most likely sum, and the chance of at least 10.
Figure (svg): The dice histogram with the tallest bar and a tail region marked
The peak is at 7. The three bars from 10 upward hold 3, 2 and 1 outcomes, totalling 6 of 36.
Worked example
Example 2, both parts.
\[ \text{From the two-dice distribution, find the most likely sum and } P(X \geq 10). \]
Find the tallest bar
Why: Six outcomes give a sum of 7, more than any other.
\[ X = 7 \]
List the values in the range
Why: Ten, 11 and 12.
Add their probabilities
Why: Three over 36 plus 2 over 36 plus 1 over 36.
\[ \frac{6}{36} \]
Simplify
Why: Six over 36.
\[ \frac{1}{6},\text{ about } 0.167 \]
Figure (svg): The dice histogram with the tallest bar and a tail region marked
\[ X = 7; \qquad P(X\geq 10) = \tfrac{1}{6} \]
Verify: check with the complement
Why: The complement is a sum of at most 9, covering 30 of the 36 pairs, so its probability is 30 over 36 — and 30 over 36 plus 6 over 36 is 1. The complement check works on a distribution exactly as it did in Lesson 10.4, since the bars partition everything.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 725-725
Fill the middle
Example 2b.
Fill in the blanks
\frac6___+\frac______+\frac______ = \frac___}___
Why: Six of the 36 pairs give a sum of at least 10, so the probability is one sixth. The three values are disjoint, so their probabilities simply add.
Worked example
Guided Practice 2.
\[ \text{For two tetrahedral dice, find the most likely sum and } P(X \leq 3). \]
Find the tallest bar
Why: Four of 16 pairs give a sum of 5.
\[ X = 5 \]
List the values at most 3
Why: Two and 3.
Add their counts
Why: One plus 2.
\[ 3\text{ of } 16 \]
Divide
Why: Three over 16.
\[ 0.1875 \]
Figure (svg): The solution to Worked example the four-sided dice shown as a ladder of expressions, one row per algebraic move
\[ X = 5; \qquad \tfrac{3}{16} = 0.1875 \]
Verify: compare with the six-sided case
Why: The peak sits at the middle of the range in both: 7 out of 2 to 12, and 5 out of 2 to 8. And the two lowest sums cover 3 of 16 here against 3 of 36 there, so the small dice make extreme sums relatively more likely — fewer pairs means less room for the middle to dominate.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 725-725
Error analysis
A student finds the probability that the sum of two dice is at least 10.
Annotate
On: \( P(X \geq 10) = P(X = 10) = \tfrac{3}{36} \)
A range covers every bar from the boundary onward. Listing the included values before adding anything makes the omission impossible.
Sorting
Read the boundary carefully.
Sort into buckets
For P(X at least 10) with two dice, sort each sum.
At least includes the boundary and more than does not, exactly as in the at-least counting problems of Lesson 10.2.
Comparison
Fill the blanks. Same shape, different sizes.
Comparison matrix
| Question | Two six-sided dice | Two four-sided dice |
|---|---|---|
| Possible sums | 2 to 12 | 2 to 8 |
| Total pairs | 36 | 16 |
| Most likely sum | 7 | 5 |
| Its probability | 6/36 = 1/6 | 4/16 = 1/4 |
The peak is more pronounced with the smaller dice, at a quarter rather than a sixth, because there are fewer values to spread the probability across.
Prediction
Commit before reasoning.
Predict first
Why are the probabilities of sums 10, 11 and 12 added rather than multiplied?
Correct: Because the sum takes exactly one value, so the three are disjoint alternatives.
\[ \text{disjoint values} \;\Longrightarrow\; \text{add their probabilities} \]
Why: A single roll produces one sum, so the events sum equals 10, sum equals 11 and sum equals 12 cannot happen together. Disjoint events add, as Lesson 10.4 established. Multiplying would answer a different question entirely — the chance of getting all three sums on three separate rolls.
Section
Section 3
Concept
A binomial experiment has n independent trials, each with only two outcomes, and the same success probability p every time. The probability of exactly k successes is the number of arrangements times the two powers.
\[ P(k) = \;_nC_k\,p^k(1-p)^{n-k} \]
Each part has a job: the combination counts which trials succeeded, and the two powers give the probability of that particular pattern.
Figure (svg): The binomial probability formula with each of its three parts labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 725-725 — Binomial Experiments
Picture it
The formula with each part explained.
Figure (svg): The binomial probability formula with each of its three parts labelled
The powers come straight from Lesson 10.5's multiplication rule; the combination is Lesson 10.2's count of which trials succeeded.
Worked example
Deriving the formula from the previous two lessons.
\[ \text{Explain each factor of } \;_nC_k\,p^k(1-p)^{n-k} \text{ for } n = 6, \; k = 2, \; p = 0.41. \]
One particular pattern
Why: Two successes and 4 failures, in a fixed order.
\[ (0.41) ^{2}(0.59) ^{4} \]
Count the patterns
Why: Which 2 of the 6 trials succeeded.
\[ 6 C 2 = 15 \]
Combine
Why: Each of the 15 patterns has the same probability.
\[ 15 \times\text{ that product} \]
Compute
Why: Fifteen times 0.1681 times 0.121174.
\[ \text{about } 0.306 \]
Figure (svg): The binomial probability formula with each of its three parts labelled
\[ \;_6C_2(0.41)^2(0.59)^4 \approx 0.306 \]
Verify: check that all the patterns are equally likely
Why: Any arrangement of 2 successes and 4 failures has the same probability, since multiplication does not care about order — 0.41 twice and 0.59 four times whichever trials they came from. That is why one product can be multiplied by the count instead of being written out fifteen times.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 725-726
Sorting
Check all three conditions.
Sort into buckets
Sort each situation.
The last fails a condition that is easy to miss: n must be fixed before the experiment starts, and rolling until something happens has no fixed n.
Worked example
Deciding whether a situation is binomial.
\[ \text{Which are binomial? Six coin tosses; six cards drawn without replacement; six free throws by one player.} \]
Coin tosses
Why: Independent, two outcomes, fixed probability.
Cards without replacement
Why: The trials are not independent.
Free throws
Why: Independent, made or missed, one player's fixed rate.
Name the failing condition
Why: Removing a card changes the next probability.
Figure (svg): The solution to Worked example test the three conditions shown as a ladder of expressions, one row per algebraic move
\[ \text{binomial}, \; \text{not}, \; \text{binomial} \]
Verify: check what would fix the second
Why: Drawing the six cards WITH replacement would restore independence and a fixed probability, making it binomial. So the situation is not inherently unsuitable — the sampling method is what breaks the conditions, which is the same replacement question as in Lesson 10.5.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 725-725
Trap
\[ P(2 \text{ of } 6) = (0.41)^2(0.59)^4 \]
Multiply the probabilities of the trials
Why: One particular pattern is computed and reported.
\[ \approx 0.0204 \quad \text{(wrong)} \]
That is the probability of one SPECIFIC pattern, such as the first two succeeding. There are 15 different patterns with two successes.
\[ \;_6C_2(0.41)^2(0.59)^4 = 15(0.0204) \approx 0.306 \]
Multiply by the number of arrangements
Why: Any two of the six trials could be the successful ones.
\[ \;_6C_2 = 15 \]
The combination is easy to omit because the powers look like a complete answer. Checking that the whole distribution totals 1 catches the omission immediately.
Fill the middle
Example 3.
Fill in the blanks
P(k=2) = \;_6C_2(0.41)^2(0.59)^4, \quad \;_6C_2 = 15
Why: Six choose 2 is 15, the number of ways two of the six households could be the owners. Without that factor the answer would be fifteen times too small.
Matching
Three parts, three roles.
Match the pairs
Why: The last row is the check the book recommends: adding every probability in a binomial distribution must give 1, which catches a dropped combination or a mistyped power at once.
Prediction
Commit before reasoning.
Predict first
Why is the probability of exactly 2 successes in 6 trials fifteen times the probability of one particular pattern?
Correct: Because there are 15 different patterns with 2 successes, each equally likely and mutually exclusive.
\[ 15 \text{ equal, disjoint patterns} \;\Longrightarrow\; 15 \times \text{one of them} \]
Why: Each pattern has probability p squared times the failure probability to the fourth, and no two patterns can happen at once, so their probabilities add — fifteen equal terms, which is fifteen times one of them. The combination is doing the counting and the powers the probability, exactly the split that the last two lessons prepared.
Section
Section 4
Concept
Apply the formula for k equal to 0 up to n and the whole distribution follows. Displayed as a histogram it shows which outcome is most likely and how quickly the probabilities fall away.
\[ P(k) = \;_6C_k(0.41)^k(0.59)^{6-k} \]
The peak sits near n times p, so it can be predicted before any arithmetic — a useful check on the completed table.
Figure (svg): A binomial histogram for six households surveyed at a 41 percent rate
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 726-726 — Construct a binomial distribution
Picture it
Example 3: six households at a 41 percent ownership rate.
Figure (svg): A binomial histogram for six households surveyed at a 41 percent rate
The tallest bar is at 2, with 3 close behind — and n times p is 2.46, sitting between them exactly as expected.
Worked example
Example 3.
\[ \text{With } n = 6 \text{ and } p = 0.41, \text{ find } P(k) \text{ for every } k. \]
k equal to 0
Why: One arrangement, six failures.
\[ 0.59 ^{6}\text{ about } 0.042 \]
k equal to 1 and 2
Why: Six and 15 arrangements.
\[ \text{about } 0.176\text{ and } 0.306 \]
k equal to 3 and 4
Why: Twenty and 15 arrangements.
\[ \text{about } 0.283\text{ and } 0.148 \]
k equal to 5 and 6
Why: Six and one arrangement.
\[ \text{about } 0.041\text{ and } 0.005 \]
Figure (svg): A binomial histogram for six households surveyed at a 41 percent rate
\[ P(2) \approx 0.306 \text{ is the largest} \]
Verify: add the seven probabilities
Why: They total 1.001, which is 1 to within rounding. The book recommends exactly this check, and it catches a dropped combination or a mistyped power straight away — a missing factor of 15 would leave the total far short of 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 726-726
Fill the middle
Example 3.
Fill in the blanks
P(k=2) = \;_6C_2(0.41)^2(0.59)^4}
Why: Six minus 2 is 4, so four households do not own a ball. The two exponents always total n, since every trial has one outcome or the other.
Worked example
Example 4 and Guided Practice 3 and 4.
\[ \text{Find the most likely outcome and } P(k \leq 2), \text{ for } p = 0.41 \text{ and again for } p = 0.61. \]
p equal to 0.41: find the peak
Why: The largest probability is at k equal to 2.
\[ 2\text{ households} \]
p equal to 0.41: add three bars
Why: Point three zero six plus 0.176 plus 0.042.
\[ \text{about } 0.524 \]
p equal to 0.61: find the peak
Why: Now n times p is 3.66, and the peak is at 4.
\[ 4\text{ households} \]
p equal to 0.61: add three bars
Why: About 0.129 plus 0.033 plus 0.004.
\[ \text{about } 0.166 \]
Figure (svg): A binomial histogram for six households surveyed at a 41 percent rate
\[ k = 2, \; 0.524; \qquad k = 4, \; 0.166 \]
Verify: check both peaks against n times p
Why: Six times 0.41 is 2.46 and the peak is at 2; six times 0.61 is 3.66 and the peak is at 4. In both cases the peak is the whole number nearest n times p, which is a reliable prediction and a quick check on a completed table.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 726-726
Error analysis
A student computes the probability that 2 of 6 households own a soccer ball.
Annotate
On: \( \;_6C_2(0.41)^2(0.59)^2 \approx 0.878 \)
The two exponents must total n, since every trial is either a success or a failure. Checking that they add to n catches this instantly.
Ranking
For n equal to 6 and p equal to 0.41, smallest first.
Put in order
Why: Notice that P of 0 and P of 5 are almost equal at 0.042 and 0.041 even though they sit at opposite ends. With p below one half the distribution leans left, so the low end is not as unlikely as symmetry would suggest.
Comparison
Fill the blanks. Same six households, different p.
Comparison matrix
| Quantity | p = 0.41 | p = 0.61 |
|---|---|---|
| n times p | 2.46 | 3.66 |
| Most likely k | 2 | 4 |
| P(at most 2) | about 0.524 | about 0.166 |
| Which way it leans | toward the low end | toward the high end |
Raising p from 0.41 to 0.61 moves the whole distribution to the right and cuts the chance of at most two owners from over a half to about one sixth.
Prediction
Commit before reasoning.
Predict first
For n equal to 20 trials with p equal to 0.3, where would you expect the tallest bar?
Correct: Near k equal to 6, which is n times p.
\[ np = 20 \times 0.3 = 6 \]
Why: Twenty trials with a 30 percent success rate give about 6 successes on average, and the distribution peaks there. That product is the distribution's mean, and predicting it before computing anything is a strong check: a table peaking at 15 would signal that p and 1 minus p had been swapped.
Section
Section 5
Concept
A distribution is symmetric when its left and right halves mirror each other, and skewed when one tail is much longer than the other. A binomial distribution is symmetric exactly when p is one half.
\[ p = 0.5 \;\Longrightarrow\; \text{symmetric} \]
For p above one half the bars pile up on the right with a long left tail, and for p below one half the reverse.
Figure (svg): Two binomial histograms for eight trials, one symmetric and one skewed
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 727-727 — Classify distributions as symmetric or skewed
Picture it
Example 5: p equal to 0.5 and p equal to 0.9.
Figure (svg): Two binomial histograms for eight trials, one symmetric and one skewed
The first is a mirror image about 4; the second is crowded against 8 with almost nothing below 4.
Worked example
Example 5, both parts.
\[ \text{Describe the shape of the binomial distribution for } n = 8 \text{ with } p = 0.5 \text{ and with } p = 0.9. \]
p equal to 0.5: find the peak
Why: Eight times one half is 4, the middle value.
\[ \text{peak at } 4 \]
p equal to 0.5: compare the halves
Why: P of 0 equals P of 8, P of 1 equals P of 7, and so on.
p equal to 0.9: find the peak
Why: Eight times 0.9 is 7.2.
\[ \text{peak near } 7\text{ or } 8 \]
p equal to 0.9: compare the tails
Why: Almost nothing below 4, but a long stretch of tiny bars.
Figure (svg): Two binomial histograms for eight trials, one symmetric and one skewed
\[ p = 0.5: \text{ symmetric}; \quad p = 0.9: \text{ skewed} \]
Verify: explain the symmetry algebraically
Why: With p equal to one half, both powers become one half to the n whatever k is, so the probability depends only on the combination — and n choose k equals n choose n minus k. That symmetry of Pascal's triangle is exactly the symmetry of the histogram, and it fails for any other p.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 727-727
Sorting
Only p equal to one half is symmetric.
Sort into buckets
Sort each binomial distribution.
The value of n changes how many bars there are but never whether the distribution is symmetric — that depends on p alone.
Worked example
Comparing the two soccer distributions.
\[ \text{Describe the shapes for } n = 6 \text{ with } p = 0.41 \text{ and with } p = 0.61. \]
p equal to 0.41: find the peak
Why: Six times 0.41 is 2.46, below the middle of 3.
\[ \text{peaks at } 2 \]
p equal to 0.41: the longer tail
Why: More room above the peak than below.
p equal to 0.61: find the peak
Why: Six times 0.61 is 3.66, above the middle.
\[ \text{peaks at } 4 \]
p equal to 0.61: the longer tail
Why: More room below the peak than above.
Figure (svg): The solution to Worked example which way does it lean shown as a ladder of expressions, one row per algebraic move
\[ 0.41: \text{ peak low}; \quad 0.61: \text{ peak high} \]
Verify: notice the mirror relationship
Why: The two values 0.41 and 0.61 are not quite mirror images of each other about one half, but 0.39 and 0.61 would be — and their distributions would be exact reflections, with P of k for one equalling P of 6 minus k for the other. Swapping p for 1 minus p always reverses the histogram.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 726-727
Trap
\[ n = 8, \; p = 0.9 \]
Call it symmetric because it has one peak
Why: A single hump is taken as the definition of symmetry.
\[ \text{symmetric} \quad \text{(wrong)} \]
P of 8 is about 0.43 while P of 0 is under a millionth. The two ends are nothing like each other.
\[ \text{compare } P(k) \text{ with } P(n-k) \]
Test whether the halves mirror each other
Why: Symmetry means matching values at matching distances from the centre.
\[ p = 0.5 \text{ only} \]
Almost every distribution in practice has a single peak. Having one says nothing about symmetry, which is a statement about the two tails.
Fill the middle
Example 5b.
Fill in the blanks
n = 8, \; p = 0.9 \;\Longrightarrow\; np = 7.2
Why: Eight times 0.9 is 7.2, so the peak sits at 7, with 8 nearly as tall. A peak that high leaves almost no room on the right and a long stretch on the left.
Comparison
Fill the blanks. Same n, different p.
Comparison matrix
| Feature | p = 0.5 | p = 0.9 |
|---|---|---|
| Peak location | k = 4, the middle | k = 7, near the top |
| P(0) against P(8) | equal, both about 0.004 | wildly different |
| Shape | symmetric | skewed |
| Longer tail | neither | the left one |
A binomial distribution's shape is decided by p alone, and the two tails are equal only when successes and failures are equally likely.
Prediction
Commit before reasoning.
Predict first
How do the distributions for p equal to 0.3 and p equal to 0.7 compare, with the same n?
Correct: Each is the mirror image of the other, reversed left to right.
\[ P_{p}(k) = P_{1-p}(n-k) \]
Why: Swapping p for 1 minus p exchanges success with failure, so k successes in one corresponds to n minus k in the other — and n choose k equals n choose n minus k, so the probabilities match exactly. Reflecting one histogram gives the other. It also explains the symmetric case: when p equals 1 minus p, the distribution is its own mirror image.
Comparison
Fill the blanks. One table, one formula, one shape.
Comparison matrix
| Idea | What it gives | Check |
|---|---|---|
| Probability distribution | a probability for every value | they must total 1 |
| Binomial formula | P(k) = nCk p^k (1 - p)^(n - k) | the exponents must total n |
| The peak | the most likely number of successes | near n times p |
| The shape | symmetric or skewed | symmetric only when p is one half |
The three checks in the last column cost seconds each and between them catch almost every error a binomial calculation can make.
Pattern
Check the conditions, then apply the formula.
The peak sits near n times p, which is worth predicting before computing anything.
OpenStax Algebra and Trigonometry 2e, §13.6 Binomial Theorem §13.6
Check
Add the bars in the range.
Check your understanding
Rolling two dice, what is the probability that the sum is at least 10?
Answer: A
Why: The sums 10, 11 and 12 account for 3 plus 2 plus 1 of the 36 pairs.
Check
The formula has three factors.
Check your understanding
With n = 6 and p = 0.41, what is P(exactly 2 successes)?
Answer: A
Why: It is 15 times 0.41 squared times 0.59 to the fourth.
Check
Shape depends on p alone.
Check your understanding
For which binomial distribution is the histogram symmetric?
Answer: A
Why: Only p equal to one half makes the two powers equal, giving a mirror image.
Real world
A drug trial gives a new treatment to 20 patients. The old treatment works 40 percent of the time, and the researchers will call the new one promising if at least 13 of the 20 improve.
Discussion prompt
If the new treatment is really no better, find the probability of at least 13 improvements, and say what that number means for the trial.
Hint: This is binomial with n equal to 20 and p equal to 0.4.
Answer:
\[ np = 20 \times 0.4 = 8, \text{ so the peak is near } 8 \]
\[ P(k \geq 13) = \sum_{k=13}^{20} \;_{20}C_k(0.4)^k(0.6)^{20-k} \approx 0.021 \]
About 2 percent. So if the new treatment is no better than the old, a result this good would happen by chance only about one time in fifty.
That number is the foundation of statistical testing: researchers fix a threshold, often 5 percent, and treat a result rarer than that as evidence the treatment really works. It is not proof — one trial in fifty will clear the bar by luck alone, which is why results are replicated and why running many trials and reporting only the successful one is so misleading. Everything here comes from the binomial formula: assume nothing has changed, compute how surprising the observed result would be, and let that probability decide.
Commit first
Answer, then rate your confidence honestly.
Predict first
Rolling two dice, are all eleven possible sums equally likely?
Correct: No — 7 occurs in six ways and 2 in only one, so the probabilities range from 1/36 to 1/6.
\[ P(7) = \tfrac{6}{36}; \quad P(2) = \tfrac{1}{36} \]
Why: The equally likely outcomes are the 36 ordered pairs, not the 11 sums. Six pairs give a total of 7 and only one gives 2, so 7 is six times as likely. Listing possible values and assuming they are equally likely is the commonest error in building a distribution, and the total is the giveaway: eleven values at one eleventh each would sum to 1 by accident, so even the sum check does not catch it — only asking what is genuinely equally likely does.
Explain it
They can find the probability of one outcome but have never seen a distribution.
Discussion prompt
In four sentences or fewer, explain what a probability distribution is and why it is useful.
Hint: Describe the whole list at once.
Answer:
Instead of asking about one result, you list every result that could happen and write its probability next to it. Drawn as a bar chart, the tallest bar is the most likely result.
It is useful because any question you might ask can be answered from the list — add up the bars you care about. And the whole list always adds to 1, which is a free way to check you have not missed anything.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For outcomes, ask whether every item on your list is genuinely as likely as every other. For the combination, remember it counts which trials succeeded. For exponents, check that the two add to n. For shape, compare the first and last bars rather than counting peaks.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a distributions page. Top: make the full table for the sum of two dice and draw its histogram, then check that the counts total 36. Beside it, do the same for two four-sided dice. Middle: write the binomial formula with each of its three factors labelled by what it does, and note the two conditions that make a situation binomial. Bottom left: build the whole distribution for six households at 41 percent, showing at least three of the seven computations in full, and add all seven probabilities as a check. Bottom right: sketch the two shapes for eight trials at p equal to one half and at p equal to 0.9, mark n times p on each, and write one sentence explaining why only the first is symmetric.
If your seven binomial probabilities do not total 1, look first for a missing combination: the powers alone always sum to far less than 1.
Recap
Five things, and Chapter 10 is complete.
| If you see | Then |
|---|---|
| A random variable | List every value with its probability |
| A distribution | Its probabilities must total 1 |
| Fixed independent trials with two outcomes | A binomial experiment |
| Exactly k successes | nCk times p^k times (1 - p)^(n - k) |
| A range of values | Add the bars it covers |
| p equal to one half | The distribution is symmetric |
That closes Chapter 10. Chapter 11 turns to sequences and series, where terms follow a rule and their sums have closed formulas.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.6 Construct and Interpret Binomial Distributions §10.6, pp. 724-731 — everything on these slides traces back here
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