GMAT Probability

Define the event and sample space

Probability measures how likely an event is under a stated model. When outcomes are equally likely, probability = favorable outcomes / total outcomes. The chance of drawing a red marble from a bag with 3 red and 5 blue marbles is 3/8. This ratio works because each marble is equally likely to be drawn.

Before calculating, state exactly what counts as success and identify the full sample space. If a question asks for the probability that a number is even when rolling a fair six-sided die, the favorable outcomes are 2, 4 and 6 out of 6, giving 1/2. If the die is not fair, the equal-likelihood assumption no longer holds.

Probability lies from 0 to 1, or from 0 percent to 100 percent. An impossible event has probability 0 and a certain event has probability 1. Use this range as a check. A result of 1.25 usually signals that favorable outcomes were counted twice or that the denominator is wrong.

Use the complement for 'at least one'

The complement of an event is the event not happening. P(A) = 1 - P(not A). For 'at least one' questions, count the opposite case, often 'none,' then subtract from 1. This avoids listing many cases with one success, two successes and so on.

A fair coin is tossed three times. The probability of at least one head is 1 minus the probability of no heads. No heads means tails on all three tosses, probability (1/2)^3 = 1/8. Therefore at least one head has probability 7/8.

A common wrong answer is 3/2 from adding a 1/2 chance of heads for each toss. Those events can overlap: several tosses may produce heads, so adding their probabilities double-counts outcomes. The complement counts a single disjoint case.

Independent events

Events are independent when the outcome of one does not change the probability of another. For independent events, the probability that both occur is the product of their probabilities. If a fair coin lands heads and a fair die shows an even number, the probability of both is 1/2 x 1/2 = 1/4.

Independence comes from the process, not from the events having different names. Two draws with replacement from a bag are independent because the first item is returned before the second draw. Two draws without replacement are dependent because the first draw changes the contents of the bag.

For at least one success across independent repeated trials, the complement method gives 1 - (probability of failure on one trial) to the power of the number of trials. If a device has a 0.9 chance of working on each independent check, the chance it works at least once in two checks is 1 - (0.1)^2 = 0.99.

Dependent events and no replacement

When drawing without replacement, update both favorable and total counts after the first draw. If a bag has 4 red and 6 blue marbles, the probability of drawing two red marbles without replacement is 4/10 x 3/9 = 12/90 = 2/15. After the first red is removed, only three red marbles remain among nine total.

For drawing one red and one blue in either order, add the two disjoint sequences: red then blue has probability 4/10 x 6/9 = 4/15; blue then red has probability 6/10 x 4/9 = 4/15. Total probability is 8/15. The order cases do not overlap, so their probabilities can be added.

An alternative is combinations: choose one red from four and one blue from six, then divide by the number of ways to choose two marbles from ten. That is (4 x 6) / C(10,2) = 24/45 = 8/15. Use whichever method is clearer; sequential probability is often easier when the number of draws is small.

Conditional probability

Conditional probability asks for the chance of A given that B has occurred. The condition narrows the sample space to outcomes where B is true. In a class of 30 students, 18 study economics and 12 study both economics and statistics. Given that a student studies economics, the probability that the student also studies statistics is 12/18 = 2/3.

The denominator is the conditioned group, 18 economics students, not all 30 students. The unconditional probability of studying both is 12/30 = 2/5. These are different questions. Look for phrases such as 'given that,' 'among those who' or 'if it is known that.'

A conditional question can also ask whether events are independent. A and B are independent when knowing B does not change the probability of A. Compare P(A given B) with P(A). If they differ, the events are dependent.

Counting outcomes without double counting

For a small sample space, list outcomes systematically. For larger spaces, use multiplication for sequential choices and combinations when order does not matter. If a four-digit code allows digits 0 through 9 with repetition, there are 10 choices for each position and 10^4 total codes. If repetition is prohibited, there are 10 x 9 x 8 x 7 choices.

Order is central. Arranging three people in a line creates different outcomes when positions change. Choosing a three-person committee does not. If order matters, count permutations; if it does not, use combinations or a direct counting argument.

Check whether outcomes are equally likely before using favorable over total. A randomly chosen person from a group gives equal probability to each person if selection is uniform. A randomly selected number from a weighted process may not give equal probability to each possible result. The denominator is a count only when the counted outcomes have equal likelihood.

Original worked examples

At least one success

A box contains 5 working and 2 defective bulbs. Two bulbs are selected without replacement. What is the probability that at least one is defective? The complement is selecting two working bulbs: 5/7 x 4/6 = 20/42 = 10/21. Therefore the answer is 1 - 10/21 = 11/21.

A wrong approach adds the chance the first bulb is defective to the chance the second is defective, which double-counts the case where both are defective. The complement counts only the single case of no defective bulbs. Because the first draw is not replaced, the probability on the second draw is 4/6 for two working bulbs.

Conditional selection

A jar has 4 green and 6 yellow tokens. One token is drawn and set aside. Given that the first token was green, what is the probability the second token is green? There are now 3 green tokens among 9 total, so the answer is 1/3. The condition changes the denominator and numerator from the original jar.

If the first token were returned, the draws would be independent and the probability would remain 4/10 = 2/5. The words 'set aside' and 'returned' determine the model. Do not use independence by default.

Combine routes

A traveler can take one of 3 buses to a station and one of 4 trains onward. If any bus can connect to any train and choices are made independently, there are 3 x 4 = 12 routes. If the traveler chooses one route uniformly, the probability of using a particular bus is 4/12 = 1/3 because four routes use it.

This example distinguishes a route count from an outcome probability. A bus is not one twelfth of all routes; it appears in four of them. First count complete outcomes, then count how many satisfy the condition.

Common probability errors

• Adding probabilities for overlapping events without subtracting their overlap.
• Multiplying probabilities when events are dependent.
• Forgetting that no replacement changes the next draw.
• Using the full population as the denominator after a condition narrows the group.
• Counting order when the question treats selections as the same group.
• Assuming equally likely outcomes without support.
• Confusing 'at least one' with 'exactly one.'

For the union of two events, P(A or B) = P(A) + P(B) - P(A and B). If the events are mutually exclusive, their overlap is zero and simple addition works. For 'exactly one' of two events, count A without B and B without A. For 'at least one,' the complement is often shorter.

How to practice probability

Translate language into event notation or a small diagram. Mark whether events are independent, whether items are replaced and whether order matters. Work with fractions until the final step. A tree diagram is helpful for two or three sequential draws; a table can organize mutually exclusive cases.

In Quant, calculate by hand and simplify fractions. For a small number of combinations, list the cases rather than forcing a formula. In Data Insights, the calculator may help with arithmetic, but it cannot resolve whether cases overlap or whether the second draw is conditional.

When reviewing an error, state the event and its complement. If you multiplied probabilities, explain why the events are independent. If you added, check for overlap. If the denominator changed, say what is now known. Then solve a new problem with different counts so the method is tested rather than memorized.