GMAT · Counting, probability and statistical arithmetic · TOPIC LESSON
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
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01 / UNDERSTAND & APPLY
Add counts for mutually exclusive alternatives that cannot occur together. Multiply counts for successive choices when each stage has the stated number of available options for every earlier choice. If the number of options depends on the earlier choice, split into cases first.
A counting model must specify whether repetitions are allowed, whether order matters and whether every arrangement is valid. A product that ignores restrictions can count impossible outcomes; a sum that combines overlapping cases can count the same outcome twice.
ORIGINAL WORKED EXAMPLE
A cafe offers 3 soups and 4 sandwiches. A customer chooses either one soup or one sandwich. How many choices? How many meals if the customer instead chooses one of each?
Answer 7 single-item choices; 12 two-item meals.
ORIGINAL PRACTICE · 1
15.
For each of 3 letters there are 5 digit choices, so 3×5=15. Adding 3+5 counts individual symbols, not complete two-symbol codes.
02 / UNDERSTAND & APPLY
Selecting r distinct items in order from n gives n(n−1)…(n−r+1)=n!/(n−r)!. Selecting an unordered group counts each group r! times in that ordered total, so divide by r! to obtain n choose r. Factorial n! is the product 1 through n, and 0!=1.
Use roles or positions to decide whether order matters. Choosing a president and secretary differs from choosing a two-person committee. Formula selection should follow the outcome definition, not whether the wording happens to say “choose.”
ORIGINAL WORKED EXAMPLE
From 6 people, how many ways are there to assign a president and a secretary, with different people in the two roles? How many two-person committees?
Answer 30 role assignments; 15 committees.
ORIGINAL PRACTICE · 1
35.
Ordered choices give 7×6×5=210. Each committee has 3!=6 internal orderings, so 210/6=35. Leaving the total at 210 treats the same three people as six different committees.
03 / UNDERSTAND & APPLY
If n objects include groups of identical objects of sizes a,b,…, the number of distinct linear arrangements is n!/(a!b!…). Start by pretending all objects are labeled, then divide out rearrangements of identical copies that produce no visible change.
Use this correction only for genuinely indistinguishable objects. Two different people with the same initial remain distinct people. Conversely, repeated digits in a code are identical symbols even if you imagine labeling their copies while counting.
ORIGINAL WORKED EXAMPLE
How many distinct arrangements of the letters in LEVEL are possible?
Answer 30.
ORIGINAL PRACTICE · 1
10.
Choose the two positions occupied by B: 5 choose 2=10. Equivalently, 5!/(3!2!)=10. Dividing by 3 and 2 rather than their factorials gives the wrong duplicate correction.
04 / UNDERSTAND & APPLY
Complementary counting is effective when “at least one” has a complicated direct case split but “none” is simple. The desired count equals total possible outcomes minus the undesired count, provided both are counted in the same outcome space.
The complement must be exact. The complement of “at least two” is “zero or one,” not merely zero. A probability complement similarly satisfies P(event)=1−P(not event), without requiring independence.
ORIGINAL WORKED EXAMPLE
A four-digit PIN uses digits 0 through 9, with repetition allowed and leading zero allowed. How many PINs contain at least one 7?
Answer 3,439.
ORIGINAL PRACTICE · 1
46.
There are 8 choose 3=56 committees total and 5 choose 3=10 with no designers. Subtract to get 46. Counting exactly one designer omits committees with two or three designers.
05 / UNDERSTAND & APPLY
Probability equals favorable outcome count divided by total outcome count only when the counted elementary outcomes are equally likely. For two fair dice, the 36 ordered face pairs are equally likely; the eleven possible sums are not equally likely.
Specify the random mechanism. Selecting a person uniformly is different from selecting a group uniformly and then selecting a person within that group if group sizes differ. Counts alone do not define probabilities without a sampling rule.
ORIGINAL WORKED EXAMPLE
Two independent fair six-sided dice are rolled. What is the probability that their sum is 8?
Answer 5/36.
ORIGINAL PRACTICE · 1
1/2.
The prime faces are 2,3,5: three of six equally likely faces. The number 1 is not prime, so including it would overcount.
06 / UNDERSTAND & APPLY
The union A or B includes A only, B only and both. Thus P(A∪B)=P(A)+P(B)−P(A∩B). Subtract the overlap once because it appears in both event totals. This rule does not assume independence.
The intersection probability is not generally the product of the individual probabilities. Use a supplied overlap, a count of joint outcomes or a conditional probability. Also check that a probability stays between 0 and 1.
ORIGINAL WORKED EXAMPLE
A person is selected uniformly from 50 people. Twenty speak language A, 18 speak B and 8 speak both. What is the probability the person speaks at least one of the two?
Answer 3/5.
ORIGINAL PRACTICE · 1
0.7.
Use 0.4+0.5−0.2=0.7. Adding without subtracting would count the overlap twice; multiplying 0.4 and 0.5 answers an intersection only under independence, which is not the requested union.
07 / UNDERSTAND & APPLY
Mutually exclusive events cannot occur together, so their intersection has probability zero. Independent events satisfy P(A∩B)=P(A)P(B); learning one occurred does not change the probability of the other. Two mutually exclusive events with positive probabilities cannot be independent.
Independence is about probability, not a casual claim that events are unrelated. Sampling without replacement usually creates dependence because the first draw changes the composition. Mutually exclusive events still add directly for a union even though they are not independent.
ORIGINAL WORKED EXAMPLE
For a fair die, let A mean an even result and B mean a result greater than 3. Are A and B independent?
Answer No; they are neither independent nor mutually exclusive.
ORIGINAL PRACTICE · 1
1/4.
Independence permits multiplication: (1/2)(1/2)=1/4. These events can occur together, so they are not mutually exclusive; treating “heads” and “tails” as exclusive ignores that they refer to different tosses.
08 / UNDERSTAND & APPLY
P(A given B)=P(A∩B)/P(B), provided P(B)>0. Conditioning restricts the outcome space to B; the numerator is the part of B that also satisfies A. Reversing the condition usually changes the denominator and the answer.
In a table, locate the conditioned row or column first. Do not divide by the entire population after the question has told you the selected person belongs to a subgroup. Conditional probability does not itself imply a causal relationship.
ORIGINAL WORKED EXAMPLE
Among 100 people, 40 have a subscription and 24 of those subscribers attended an event. A subscriber is selected uniformly. What is the probability that person attended?
Answer 3/5.
ORIGINAL PRACTICE · 1
2/3.
The conditioned equally likely outcomes are 2,4,6. Two, 4 and 6, exceed 3, so probability is 2/3. Using 2/6 ignores the given information.
09 / UNDERSTAND & APPLY
With replacement and independent uniform draws, the composition is restored before each draw, so the same fraction may be multiplied repeatedly. Without replacement, both the total and possibly the favorable count change after a draw. Use conditional probabilities along each ordered path.
For an unordered event such as one red and one blue, count both orders or use combinations consistently. Multiplying one red-then-blue path alone misses the blue-then-red path. Do not count orders again after using unordered combinations.
ORIGINAL WORKED EXAMPLE
A bag contains 3 red and 2 blue counters. Two are drawn uniformly without replacement. What is the probability both are red? What if the first is replaced and the draws are independent?
Answer Without replacement 3/10; with replacement 9/25.
ORIGINAL PRACTICE · 1
3/5.
Red then blue has probability (3/5)(2/4)=3/10; blue then red has (2/5)(3/4)=3/10. Add disjoint orders to get 3/5. Alternatively favorable unordered pairs number 3×2=6 of 5 choose 2=10.
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