Probability and Statistics — Free Algebra 2 Review Games.

5! = 5*4*3*2*1 = 120.

4! = 4*3*2*1 = 24.

Combinations count selections without regard to order.

C(5,2) = 5!/(2!*3!) = 120/(2*6) = 10.

P(3,3) = 3! = 6.

P(5,3) = 5!/(5-3)! = 120/2 = 60.

C(3,2)*(1/2)^3 = 3/8.

The empirical rule: about 68% of data falls within 1 standard deviation.

C(6,3) = 6!/(3!*3!) = 720/(6*6) = 20.

Even: {2,4,6}. Greater than 4: {5,6}. Union: {2,4,5,6} = 4 outcomes. P = 4/6 = 2/3.

Mean = np = 10 * 0.3 = 3.

C(8,3) = 8!/(3!*5!) = 56.

A negative z-score means below the mean by that many standard deviations.

P(10,3) = 10*9*8 = 720.

\(SD = \sqrt{np(1-p)} = \sqrt{20 \cdot 0.5 \cdot 0.5} = \sqrt{5}\).

By definition, \(0! = 1\) so that formulas like \(\frac{n!}{n!}\) remain consistent when \(n=0\). The choice "\(0\)" is wrong because factorial is never zero for any nonnegative integer input. Remembering \(0!=1\) is essential because it appears constantly in permutation and combination formulas.

\(P(n,r) = \frac{n!}{(n-r)!}\) works because it counts ordered selections by removing the arrangements of the unused \((n-r)\) items. The option \(\frac{n!}{r!}\) is wrong because dividing by \(r!\) instead removes ordering among the chosen items, which is actually the combination formula. Distinguishing permutations from combinations always comes down to whether order matters.

\(C(n,r) = \frac{n!}{r!(n-r)!}\) is correct because it takes the permutation count and divides by \(r!\) to remove overcounting from order within the selected group. The choice \(\frac{n!}{(n-r)!}\) is wrong because that is the permutation formula, which still counts order. Always check whether the problem cares about order to pick the correct formula.

A trial is one instance of the random experiment, such as one coin flip, and each trial must have only two outcomes for the binomial model to apply. The choice "the total number of experiments run" is wrong because that describes \(n\), the number of trials, not a single trial itself. Recognizing individual trials is the foundation for setting up any binomial probability calculation.

The mean of a binomial distribution is \(\mu = np\) because it represents the expected number of successes across \(n\) independent trials each with success probability \(p\). The choice \(np(1-p)\) is wrong because that expression actually gives the variance, not the mean. Keeping mean and variance formulas separate is critical when analyzing binomial data.

A normal distribution produces a bell-shaped, symmetric curve centered at the mean because data values cluster near the average and taper off equally in both directions. "Skewed to the right" is wrong because skewed distributions are not symmetric, whereas the normal curve has identical tails on each side. Recognizing the bell shape helps students quickly identify when the empirical rule and z-scores can be applied.

\(C(4,0) = \frac{4!}{0!(4-0)!} = \frac{24}{1 \cdot 24} = 1\) because there is exactly one way to choose nothing from a group. The choice "\(0\)" is wrong because choosing zero items is still a valid, single outcome, not an impossible one. Remember that \(C(n,0)\) always equals \(1\) for any nonnegative \(n\).

\(P(4,2) = \frac{4!}{(4-2)!} = \frac{24}{2} = 12\) because there are 4 choices for the first position and 3 remaining choices for the second, giving \(4 \times 3 = 12\). The choice "\(6\)" is wrong because that is \(C(4,2)\), the unordered count, not the ordered permutation count. Always confirm whether order matters before choosing between \(P\) and \(C\).

\(\frac{6!}{3!} = \frac{720}{6} = 120\) because the \(3!\) in the denominator cancels the first three factors of \(6!\), leaving \(6 \times 5 \times 4 = 120\). The choice "\(720\)" is wrong because that is \(6!\) itself without any cancellation. Simplifying factorial ratios before multiplying often avoids unnecessary large computations.

Binomial trials are defined by two outcomes labeled success and failure, where 'success' refers to the outcome being counted in the probability calculation. "Even and odd" is wrong because that labeling only applies to specific numeric experiments like rolling a die, not the general binomial framework. Using the success/failure language keeps the binomial setup consistent across different real-world contexts.

The empirical rule is called the 68-95-99.7 rule because it states that about 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3 standard deviations of the mean. "Central limit rule" is wrong because the central limit theorem describes sampling distributions of means, not the spread of a single normal population. Memorizing these percentages lets students quickly estimate probabilities without full z-table calculations.

A z-score is calculated as \(z = \frac{x - \mu}{\sigma}\), which expresses how many standard deviations a value lies above or below the mean. "The raw value of a data point" is wrong because the raw value is \(x\) itself, before it is standardized relative to the mean and spread. Z-scores are essential because they allow comparison of values from different normal distributions on a common scale.

\(C(7,7) = \frac{7!}{7!0!} = 1\) because there is only one way to choose all 7 items from a group of 7 — you simply take everything. The choice "\(7\)" is wrong because that would be \(C(7,1)\), choosing just one item at a time, not all of them. Remember that \(C(n,n)\) always equals \(1\) regardless of the value of \(n\).

Since committee membership does not depend on order, this uses \(C(10,4) = \frac{10!}{4!6!} = 210\). The choice "\(5040\)" is wrong because that is \(P(10,4)\), which would apply only if the 4 people were assigned distinct ranked roles. Whenever a group has no internal order or ranking, combinations rather than permutations must be used.

LEVEL has 5 letters with L repeated twice and E repeated twice, so the arrangements are \(\frac{5!}{2!2!} = \frac{120}{4} = 30\). The choice "\(120\)" is wrong because that ignores the repeated letters and treats all 5 letters as distinct. Whenever a word contains repeated letters, divide the total factorial by the factorial of each repeat count.

The count is \(C(5,2) \times C(3,1) = 10 \times 3 = 30\) because the two selections are independent choices that must be multiplied together. The choice "\(15\)" is wrong because it only accounts for \(C(5,2)=10\) combined incorrectly with a single blue option instead of multiplying by all 3 blue choices. When a problem involves selecting from multiple distinct groups, multiply the combination counts from each group.

Using \(P(X=4) = C(8,4)(0.5)^4(0.5)^4 = 70 \times 0.00390625 \approx 0.273\), this reflects the peak probability near the mean of a symmetric binomial distribution. The choice "\(0.500\)" is wrong because it mistakes the overall probability of success per trial for the probability of exactly 4 successes out of 8. Binomial probability requires multiplying the combination count by both the success and failure probabilities raised to their respective powers.

\(P(X=3) = C(5,3)(0.5)^3(0.5)^2 = 10 \times 0.03125 = 0.3125\) because there are 10 distinct orderings of getting 3 heads among 5 flips, each with equal probability. The choice "\(0.5\)" is wrong because it wrongly treats the probability of exactly 3 heads as the same as the single-flip probability of heads. Always count the number of favorable orderings with the combination formula before multiplying by the individual outcome probabilities.

Using \(x = \mu + z\sigma = 100 + 2(15) = 130\), this converts the standardized z-score back into the original data scale. The choice "\(115\)" is wrong because that corresponds to only \(z=1\), one standard deviation above the mean, not two. Converting between z-scores and raw scores always requires multiplying by \(\sigma\) and adding \(\mu\).

The empirical rule states that approximately 95% of data lies within 2 standard deviations of the mean in a normal distribution. The choice "\(68\%\)" is wrong because that percentage corresponds to data within only 1 standard deviation, a narrower range. Knowing these benchmark percentages allows quick probability estimates without computing exact z-scores.

\(P(X=7) = C(10,7)(0.5)^{7}(0.5)^{3} = 120 \times 0.0009766 \approx 0.117\) since each answer is independently correct or incorrect with probability 0.5. The choice "\(0.700\)" is wrong because it incorrectly treats the fraction of questions answered correctly as a direct probability rather than computing the binomial term. Binomial probability with \(p=0.5\) still requires the combination count to weight how many arrangements achieve the target number of successes.

Since order matters when arranging trophies, this is \(P(7,4) = \frac{7!}{3!} = 840\). The choice "\(35\)" is wrong because that is \(C(7,4)\), which ignores the different possible orderings of the chosen 4 trophies. Arrangement problems on a shelf or in a line always require permutations, not combinations.

\(C(9,4) = \frac{9!}{4!5!} = 126\) because it counts the unordered ways to choose 4 items from a set of 9. The choice "\(3024\)" is wrong because that is \(P(9,4)\), the ordered count, which is much larger since it distinguishes arrangement order. Always divide out the factorial of the chosen group size when order does not matter.

\(P(X=1) = C(6,1)(0.1)^1(0.9)^5 = 6 \times 0.1 \times 0.59049 \approx 0.354\) because it accounts for all 6 possible positions where the single defect could occur. The choice "\(0.100\)" is wrong because it only reflects the probability of a defect on a single trial, ignoring the combination of positions and the probability of non-defects in the other trials. Binomial calculations always multiply the combination term by both success and failure probability powers.

The standard deviation of a binomial distribution is \(\sigma = \sqrt{np(1-p)}\) because it is the square root of the variance, \(np(1-p)\). The choice "\(np(1-p)\)" is wrong because that expression alone gives the variance, not its square root. Always remember to take the square root of the variance formula when standard deviation is requested.

\(P(8,5) = \frac{8!}{3!} = 6720\) because it counts ordered arrangements of 5 items selected from 8 distinct items. The choice "\(56\)" is wrong because that is \(C(8,5)\), the unordered selection count, which is much smaller than the permutation count. When order among selected items matters, always use the permutation formula rather than combinations.

The count is \(C(6,3) \times C(5,2) = 20 \times 10 = 200\) because the selections of men and women are independent and must be multiplied together. The choice "\(462\)" is wrong because that equals \(C(11,5)\), which incorrectly treats the men and women as one combined group instead of two separate categories. When forming a team with specific quotas from different groups, multiply the combinations from each group separately.

Since \(z = \frac{x-\mu}{\sigma}\), when \(x = \mu\) the numerator becomes zero, making \(z=0\) regardless of \(\sigma\). The choice "It depends on the standard deviation" is wrong because the numerator is always zero when \(x=\mu\), so dividing by any nonzero \(\sigma\) still gives zero. A z-score of 0 always indicates the data point is exactly at the mean of its distribution.

With \(p=0.25\) for red, \(P(X=2) = C(5,2)(0.25)^2(0.75)^3 = 10 \times 0.0625 \times 0.421875 \approx 0.264\). The choice "\(0.400\)" is wrong because it does not correctly weight the probability by the number of ways 2 successes can occur among 5 spins combined with the failure probability of the other 3 spins. Binomial probability problems always require raising both success and failure probabilities to the correct powers and multiplying by the combination term.

PEPPER has 6 letters with P repeated 3 times and E repeated 2 times, giving \(\frac{6!}{3!2!} = \frac{720}{12} = 60\) distinct arrangements. The choice "\(720\)" is wrong because that treats all 6 letters as unique, ignoring the repeated P's and E's. Whenever letters repeat in a word, divide the total factorial by the factorial of each repetition count.

Since the order in which cards are dealt does not change the hand itself, the correct count uses combinations: \(C(52,5)\). The choice "\(P(52,5)\)" is wrong because it would count each different ordering of the same 5 cards as a separate hand, which overcounts the actual number of unique hands. Card hands, lottery number selections, and similar unordered groupings always use combinations rather than permutations.

Since 60 is \(z=2\) above the mean, and the empirical rule states 95% of data lies within 2 standard deviations, the remaining 5% is split evenly in the two tails, giving about 2.5% above 60. The choice "\(16\%\)" is wrong because that percentage corresponds to the area above \(z=1\), not \(z=2\). Remembering the tail percentages tied to each z-score boundary helps quickly estimate proportions beyond specific cutoffs.

Variance is \(np(1-p) = 20 \times 0.25 \times 0.75 = 3.75\), since it measures the expected squared spread of successes around the mean. The choice "\(5\)" is wrong because that is just \(np\), the mean, not the variance formula which also requires multiplying by \((1-p)\). Distinguishing between mean \(np\) and variance \(np(1-p)\) is essential when solving multi-step binomial problems.

At least 4 made means \(P(X=4) + P(X=5) = C(5,4)(0.7)^4(0.3) + (0.7)^5 \approx 0.360 + 0.168 = 0.528\). The choice "\(0.360\)" is wrong because it only represents \(P(X=4)\) and omits the additional probability of making all 5 shots. "At least" problems always require summing multiple binomial terms rather than computing a single probability.

The count is \(C(4,2) \times C(6,3) = 6 \times 20 = 120\) since the women and men are chosen independently and multiplied together to satisfy the quota. The choice "\(252\)" is wrong because that equals \(C(10,5)\), which ignores the requirement that exactly 2 of the 5 members must be women. Quota-based committee problems always split into separate combination calculations for each subgroup before multiplying.

MISSISSIPPI has 11 letters with I repeated 4 times, S repeated 4 times, and P repeated 2 times, giving \(\frac{11!}{4!4!2!} = 34{,}650\). The choice "\(39{,}916{,}800\)" is wrong because that is \(11!\) without adjusting for any of the repeated letters. When multiple letters repeat different numbers of times, each repetition's factorial must be included in the denominator separately.

A z-score of \(-2.5\) lies far in the left tail, and standard normal tables show only about 0.6% of data falls below this point, placing it near the 0.6th percentile. The choice "about the 2.5th percentile" is wrong because that percentile corresponds to \(z=-2\), not the more extreme \(z=-2.5\). As z-scores become more negative, the corresponding percentile drops quickly toward zero, reflecting how rare such values are.

With \(p=0.25\), \(P(X=3) = C(12,3)(0.25)^3(0.75)^9 = 220 \times 0.015625 \times 0.0751 \approx 0.258\), combining the number of ways to arrange 3 correct answers with the probabilities of correct and incorrect guesses. The choice "\(0.750\)" is wrong because that is simply the probability of guessing incorrectly on a single question, not the probability across all 12 questions for exactly 3 correct. Multi-step binomial problems with more than two answer choices still require accurately identifying \(p\) before applying the formula.

Since each prize is distinct and awarded to a different contestant, order matters, giving \(P(10,4) = \frac{10!}{6!} = 5040\). The choice "\(210\)" is wrong because that is \(C(10,4)\), which would apply only if the prizes were identical and unordered. Distinct prizes assigned to different individuals always require permutations rather than combinations.

One standard deviation below and above the mean spans approximately \(6-1.897=4.1\) to \(6+1.897=7.9\), so the integers falling strictly within that range are 5, 6, and 7. The choice "\(4,5,6,7,8\)" is wrong because 4 and 8 fall outside the interval \((4.1, 7.9)\), even though they are close to the boundary. Applying the empirical rule to discrete binomial data requires rounding to the nearest whole numbers that actually fall inside the calculated interval.

A set of 3 elements has \(2^3 = 8\) total subsets, including the empty set, so removing the empty set leaves \(8 - 1 = 7\) nonempty subsets. The choice "\(8\)" is wrong because it fails to exclude the empty subset, which does not count as a nonempty group. Counting all possible subgroup sizes from a set always relies on the power set formula \(2^n\) before adjusting for any restrictions.

Using \(x = \mu + z\sigma = 80 + 1.28(10) = 92.8\), this converts the standardized cutoff for the 90th percentile into the original score scale. The choice "\(90.0\)" is wrong because it simply adds 10 to the mean without applying the correct z-value that identifies the top 10% cutoff. Identifying percentile cutoffs in a normal distribution always requires the specific z-score associated with that percentile, not a rounded estimate.

\(P(X \geq 2) = 1 - P(X=0) - P(X=1) = 1 - 0.1176 - 0.3025 \approx 0.580\), using the complement rule to avoid summing many individual terms. The choice "\(0.420\)" is wrong because it does not correctly account for both \(P(X=0)\) and \(P(X=1)\) being subtracted from 1. When a problem asks for "at least" a value, computing the complement is often far more efficient than adding all higher terms individually.

Circular arrangements are counted as \((n-1)! = 7! = 5040\) because rotating the entire arrangement produces an identical seating, so one position is fixed to eliminate duplicate rotations. The choice "\(40{,}320\)" is wrong because that is \(8!\), which counts every rotation as a separate arrangement even though they are functionally identical around a circle. Circular permutation problems always require dividing out the rotational symmetry by using \((n-1)!\) instead of \(n!\).

Dividing \(P(6,3)=120\) by \(3! = 6\) gives \(20\), which matches \(C(6,3)\) because dividing out the internal orderings of the 3 selected items converts the ordered count into an unordered one. The choice "\(40\), which equals \(C(6,2)\)" is wrong because it misapplies the division and references a combination of a different size that doesn't correspond to the calculation shown. This relationship, \(C(n,r) = \frac{P(n,r)}{r!}\), is the core link between permutations and combinations that students should carry into any selection problem.