conditional probability . data analysis . Digital SAT Math . IB IGCSE A-Level SAT prep . margin of error . percentages . ratios . scatterplots . statistics

Digital SAT Math Data Analysis Practice
Practice Drills with Desmos Split

Digital SAT Math · Problem-Solving and Data Analysis · ~15% of the Math section

Covers Ratios and percentages · One-variable data · Scatterplots and two-variable data · Probability and statistical inference
Levels Easy to Very Hard · Adaptive Module 1 and Module 2 style
Written by 25 years IB, IGCSE, and A-Level teaching experience
Format Free PDF sets + Premium Full Pack (108 questions, all five layers)
Teaching Insight

The denominator that costs the mark: why conditional probability questions have one trap and one trap only

In over 25 years of teaching students preparing for standardised mathematics exams from IB, IGCSE, and A-Level backgrounds, the conditional probability question on a two-way table is the single question type where I see the most students choose a wrong answer they are genuinely confident about. The question says: given that a student passed the exam, what is the probability that the student is in twelfth grade? The table shows 200 students total, 125 who passed, and 45 of the 125 who are in twelfth grade. The student writes 45 divided by 200, which equals 0.225, and marks it as the answer. The correct answer is 45 divided by 125, which equals 0.36. The student was not confused about probability. They understood that the answer involves the number 45. They used the wrong denominator because the phrase given that a student passed felt like context rather than a restriction, and they defaulted to the full table total. The single habit that prevents this error is pausing before writing any denominator and asking: what is the total number of people inside the condition the question has given me? If the condition is given that a student passed, that total is 125, not 200. That pause takes two seconds. These drills build it as a reflex on every conditional probability question in the set.

Recognition Training

Digital SAT Math Data Analysis — Practice Set I

easy medium

Six questions covering a direct ratio scale-up, a percentage of total calculation, converting a percent increase to a decimal multiplier, finding the mean of a small data set using symmetry as a shortcut, identifying the mode from a frequency list, and interpreting slope as a rate of change in a real-world linear model. Each question includes the algebraic method, the Desmos shortcut where applicable, a faster-path verdict, and the specific mistake students from IB and IGCSE backgrounds most commonly make.

Digital SAT Math Data Analysis — Practice Set II

medium hard

Six questions covering reading the predicted value from a line-of-best-fit equation, interpreting the y-intercept in a real-world context, computing the residual as actual minus predicted with the correct sign, basic two-way table probability using the full table total, joint probability requiring the correct cell divided by the full total, and conditional probability requiring the correct restricted denominator. The conditional probability question introduces the denominator-restriction habit explicitly with a step-by-step check.

Digital SAT Math Data Analysis — Practice Set III

hard vh

Six Hard to Very Hard questions drawn from the Module 2 difficulty ceiling: successive percent change where a 25 percent increase followed by a 20 percent decrease returns exactly to the original price, an inverse proportion problem using total-work calculation, a three-term ratio chain threaded through a shared common value, a weighted mean where group sizes must be used to compute the combined total before dividing, a without-replacement probability where the second draw denominator decreases by one, and a conditional probability with a reversed condition that changes which row becomes the restricted denominator.

The 4 Patterns Behind Every Lost Mark

01

Adding Percent Changes Instead of Multiplying the Remaining Factors

A 25 percent increase followed by a 20 percent decrease does not produce a 5 percent net increase. The correct calculation multiplies the two remaining-value factors: 1.25 times 0.80 equals 1.00, meaning the original value is exactly restored. Students who add the two percentages are applying a rule that only works when both changes apply to the same base. On the Digital SAT the second percent change always applies to the already-modified value, so the two factors must always be multiplied.

02

Using the Full Table Total as the Denominator in a Conditional Probability

When a two-way table question says given that a condition is true, the denominator becomes the total count of people inside that condition, not the full table total. If the condition is given that the student plays a sport, the denominator is the number of students who play a sport, not the number of students surveyed. Using the full table total is the single most common error on two-way table probability questions because the given-that phrase feels like context rather than a mathematical restriction on which population the probability is computed over.

03

Averaging Group Means Instead of Computing the Weighted Mean

When two groups with different sizes have different means, the combined mean is not the simple average of the two group means. It is the total points across both groups divided by the total number of people. A group of 10 with mean 70 and a group of 15 with mean 80 have a combined mean of 700 plus 1200 divided by 25, which equals 76, not 70 plus 80 divided by 2, which equals 75. Averaging the means directly only gives the correct result when both groups are exactly the same size.

04

Treating Without-Replacement Draws as Independent Events

When two items are drawn from a set without replacement, the second draw has one fewer item available than the first. If a bag contains 4 red and 6 blue marbles and two are drawn without replacement, the probability that the first is red is 4 over 10, but the probability that the second is also red is 3 over 9, not 4 over 10. Multiplying 4 over 10 by 4 over 10 gives the probability for draws with replacement. The without-replacement calculation always uses a reduced denominator on the second draw.

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