Probability of Combined Events

A common error is to say “it either happens or it doesn’t, so the probability is \( \frac{1}{2} \).” But this confuses two outcomes with two equally likely outcomes. Listing the full sample space shows four equally likely results: HH, HT, TH, TT. Only one of these is “both heads,” giving a probability of \( \frac{1}{4} \), not \( \frac{1}{2} \).

A second approach: the two flips are independent events. \( P(\text{heads on first}) = \frac{1}{2} \) and \( P(\text{heads on second}) = \frac{1}{2} \), so \( P(\text{both heads}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \). Whenever we require several independent conditions to all be met, we multiply their probabilities — which always produces a result smaller than either individual probability (when both are less than 1).

The even numbers are \( \{2, 4, 6\} \) and the numbers greater than 3 are \( \{4, 5, 6\} \). These groups overlap — the outcomes 4 and 6 belong to both. If we simply add \( \frac{3}{6} + \frac{3}{6} = 1 \), we have counted \( \{4, 6\} \) twice and arrived at a probability of 1, implying it is certain — clearly wrong, since rolling a 1 or a 3 satisfies neither condition.

The correct method subtracts the overlap: \( P(\text{even OR} > 3) = \frac{3}{6} + \frac{3}{6} – \frac{2}{6} = \frac{4}{6} = \frac{2}{3} \). The general addition rule is \( P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B) \), and the subtraction step is only skippable when \( A \) and \( B \) are mutually exclusive — that is, when they cannot overlap.

Two events are independent when knowing the outcome of one gives no information about the other. Suppose the bag contains 3 red and 2 blue balls. Initially, \( P(\text{first ball is red}) = \frac{3}{5} \). If a red ball has been removed, only 4 balls remain, of which 2 are red, so \( P(\text{second ball is red} \mid \text{first was red}) = \frac{2}{4} = \frac{1}{2} \). Since \( \frac{1}{2} \neq \frac{3}{5} \), the probability of the second draw changes depending on the first result — the definition of dependence.

Had the first ball been replaced, the bag would always contain 3 red from 5, and the events would be independent. The key distinction is whether the composition of the bag changes between draws. Without replacement, it always does.

A fair die has no memory. Each roll is a fresh, independent event — the die cannot remember what it showed before, and the six faces remain equally likely every time it is thrown. The probability of each face is always \( \frac{1}{6} \), regardless of past outcomes.

This matters because many people believe a 6 is “due” after a long absence, or “less likely” right after one has appeared. Both beliefs are wrong. Over a very large number of rolls the proportion of 6s will approach \( \frac{1}{6} \), but this is the long-run average — not a guarantee within any short sequence. Past outcomes cannot influence a fair future roll.

Example: Rolling a fair die and getting a 3; rolling the same die and getting a 5 (on a single roll). These cannot both happen at once.

Another: Drawing one card from a pack and getting a club; drawing the same card and getting a heart. A card cannot belong to two suits.

Creative: It is raining and it is simultaneously not raining at the same location. The physical impossibility makes these mutually exclusive with absolute certainty — \( P(A \text{ and } B) = 0 \).

Trap: “Coin 1 lands heads (A) and coin 2 lands tails (B) — these are mutually exclusive.” This is wrong. Both conditions can be satisfied simultaneously: the outcome, written as a coordinate pair (Coin 1, Coin 2) → (H, T), shows A and B occurring at the same time as a single compound outcome. Students often think that because each event refers to a different object, they cannot both occur. Mutual exclusivity depends on whether both events can simultaneously be true — not on whether they involve different objects.

Example: Rolling a fair die and flipping a fair coin. The die result cannot affect the coin, and vice versa.

Another: Drawing a card from a deck, recording its suit, replacing it and shuffling, then drawing again. The replacement restores the original setup, so the second draw is unaffected by the first.

Creative: Whether a volcano erupts in Iceland today and whether a randomly chosen student rolls a 6 on a die. Physical separation makes any influence impossible — these are independent by the laws of nature.

Trap: “Picking a ball from a bag, then picking a second ball from the same bag — these are independent because they are separate actions.” Wrong when there is no replacement. Without replacement, the first pick changes the contents of the bag, so the probability of the second pick depends on what was drawn first. Separate actions are not automatically independent; independence requires that the outcome of one cannot affect the probabilities of the other.

Example: Drawing one card from a standard deck. Let A = red card \( \left(P = \frac{26}{52}\right) \) and B = king \( \left(P = \frac{4}{52}\right) \). These overlap: the king of hearts and king of diamonds are both red kings. \( P(A \text{ and } B) = \frac{2}{52} \). So \( P(A \text{ or } B) = \frac{26}{52} + \frac{4}{52} – \frac{2}{52} = \frac{28}{52} = \frac{7}{13} \), not \( \frac{30}{52} \).

Another: Rolling a die, A = even number \( \{2,4,6\} \), B = multiple of 3 \( \{3,6\} \). Overlap = \( \{6\} \). \( P(A \text{ or } B) = \frac{3}{6} + \frac{2}{6} – \frac{1}{6} = \frac{4}{6} \), not \( \frac{5}{6} \).

Creative: In a class, 70% of students like pizza and 60% like pasta. If 40% like both, then \( P(\text{pizza or pasta}) = 0.7 + 0.6 – 0.4 = 0.9 \). Simply adding gives 1.3, which exceeds 1 and is impossible as a probability.

Trap: A = rolling a 1, B = rolling a 2 on a single die — \( P(A \text{ or } B) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \). This addition works because these events are mutually exclusive (cannot both occur on one roll), so there is no overlap to subtract. This is the one situation where the simple addition formula is valid — not the general case the question asks for.

Example: A bag contains 3 red and 2 blue balls. A = first ball drawn is red, B = second ball drawn is red (without replacement). \( P(A \text{ and } B) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10} \neq 0 \) (not mutually exclusive). \( P(A) \times P(B) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \neq \frac{3}{10} \) (not independent).

Another: A = a student studies for an exam, B = the student passes the exam. Both can occur at the same time (not mutually exclusive), but studying makes passing more likely — so they are dependent (not independent).

Creative: Drawing one card from a standard deck. A = card is a heart, B = card is red. \( P(A \text{ and } B) = \frac{13}{52} = \frac{1}{4} \neq 0 \) (not mutually exclusive, since all hearts are red). \( P(A) \times P(B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \neq \frac{1}{4} \) (not independent).

Trap: “Without replacement automatically means mutually exclusive.” This confuses dependence with mutual exclusivity. Dependent events can both occur — they just influence each other’s probabilities. Drawing two red balls in a row is possible; it simply becomes less likely after the first red is removed.

Example: A Venn diagram showing students who travel to school by bus and students who walk. Since they cannot be in both groups simultaneously, the intersection is empty.

Another: Rolling a die where Circle A represents rolling an even number and Circle B represents rolling a 5. These are mutually exclusive, so they don’t overlap.

An empty intersection represents mutually exclusive events. Visually, the two circles in the Venn diagram do not overlap at all, representing that $P(A \text{ and } B) = 0$.

True case: \( P(\text{rolling a 6}) = \frac{1}{6} \) and \( P(\text{flipping heads}) = \frac{1}{2} \). These are independent. \( P(\text{both}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \), which is less than both \( \frac{1}{6} \) and \( \frac{1}{2} \).

False case: Let \( P(A) = 1 \) (a certain event, e.g. “the sun rises tomorrow”) and \( P(B) = 0.4 \). These are independent. \( P(A \text{ and } B) = 1 \times 0.4 = 0.4 = P(B) \). The combined probability equals \( P(B) \) — not smaller than both. The statement fails whenever one event is certain.

True case: The statement is true in a very sneaky edge case. If one of the events is impossible (e.g., \( P(A) = 0 \)), then \( P(A \text{ and } B) = 0 \). Multiplying their probabilities gives \( 0 \times P(B) = 0 \). Because \( P(A \text{ and } B) = P(A) \times P(B) \), the events are technically independent.

False case: For any standard scenario where both events are possible, the statement is false. If A and B are mutually exclusive with \( P(A) > 0 \) and \( P(B) > 0 \), then \( P(A \text{ and } B) = 0 \). But for independence we need \( P(A \text{ and } B) = P(A) \times P(B) \), and since both probabilities are positive, \( P(A) \times P(B) > 0 \). So \( 0 \neq P(A) \times P(B) \) — the events are not independent.

In fact, mutual exclusivity is normally the strongest form of dependence: knowing A has occurred tells us with certainty that B has not.

True case: A = rolling a 3 on a die, B = rolling a 5 on a die (single roll). These are mutually exclusive. \( P(A \text{ or } B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \). Here the formula holds exactly because \( P(A \text{ and } B) = 0 \).

False case: A = rolling an even number \( \{2,4,6\} \), B = rolling a number greater than 3 \( \{4,5,6\} \). Overlap = \( \{4,6\} \). \( P(A \text{ or } B) = \frac{3}{6} + \frac{3}{6} – \frac{2}{6} = \frac{4}{6} \), but the formula would give \( \frac{3}{6} + \frac{3}{6} = 1 \). The correct general rule is \( P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B) \), and the subtraction is only zero when the events are mutually exclusive.

True case: A = rolling a 6 on a die, B = flipping heads on a coin. Both can occur simultaneously. They are independent. \( P(A \text{ and } B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} = P(A) \times P(B) \). ✓

False case: A = first ball drawn from a bag is red (without replacement), B = second ball drawn is red. Both can occur (they are not mutually exclusive). But \( P(A \text{ and } B) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10} \) while \( P(A) \times P(B) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \). These are not equal — the events are dependent. “Not mutually exclusive” does not imply “independent.” The multiplication rule \( P(A) \times P(B) \) applies only when the events are independent.

The student has confused “or” with “and” and applied the multiplication rule incorrectly. The multiplication rule \( P(A \text{ and } B) = P(A) \times P(B) \) is used when we need both events to happen — but this question asks for either one. This is a classic “and/or confusion” error.

Rolling a 3 and rolling a 5 on the same die are mutually exclusive events (you cannot get both on a single roll), so the correct rule is addition: \( P(3 \text{ or } 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \). On a single trial, “or” between mutually exclusive events means adding, not multiplying.

The answer of \( \frac{1}{2} \) is correct, but the reasoning is the “equally likely outcomes” fallacy. “At least two heads” and “fewer than two heads” are indeed complementary, but complementary events are only equally likely when they happen to be symmetric — they are not equal in general just because there are two of them.

The correct method lists all 8 equally likely outcomes (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT) and counts those with at least 2 heads: HHH, HHT, HTH, THH — that is 4 out of 8, giving \( \frac{4}{8} = \frac{1}{2} \). The student’s answer happens to be correct despite the flawed logic. To see why the reasoning is wrong, apply it to four coins: the same logic would give \( \frac{1}{2} \), but the true answer is \( \frac{11}{16} \).

The student has forgotten that some kings are red. The king of hearts and king of diamonds are both red cards and kings — they satisfy both conditions and have been counted twice. This is the “forgetting the overlap” error in the addition rule.

The correct calculation uses the general addition rule: \( P(\text{red or king}) = \frac{26}{52} + \frac{4}{52} – \frac{2}{52} = \frac{28}{52} = \frac{7}{13} \). The subtraction of \( \frac{2}{52} \) removes the double-counting of the two red kings. The formula \( P(A) + P(B) \) with no subtraction is only valid for mutually exclusive events — and red cards and kings clearly overlap.

The student has applied the addition rule when the multiplication rule is needed. This is the “adding instead of multiplying for AND events” error. Adding probabilities finds \( P(A \text{ or } B) \) — the chance of rain on Monday or Tuesday (or both). The question asks for rain on Monday and Tuesday, which requires both conditions to be met simultaneously.

Since the two days are independent, the correct calculation is \( P(A \text{ and } B) = P(A) \times P(B) = 0.4 \times 0.3 = 0.12 \). A useful check: the probability of two events both happening should be lower than either individual probability. Getting 0.7 — which is larger than both 0.4 and 0.3 — should immediately signal that something has gone wrong.

Students often confuse when to add and when to multiply on a tree diagram. You multiply along the branches (AND) because the events are sequential and combined. You add down the columns (OR) when combining mutually exclusive paths.

The student incorrectly added along the branches. The correct calculation is \( 0.7 \times 0.3 = 0.21 \). Getting a probability of \( 1.0 \) (absolute certainty) for just one specific outcome out of four possible outcomes is a clear signal that the operation was incorrect.