Chapter 13.2 — Exercise 13.1 — Complementary Events
Probability of complementary events. This is Lesson 2 of 3 in Chapter 13: Probability.
The Complement Rule in Practice
Exercise 13.1 works the complementary-event relationship P(E) + P(not E) = 1 from every direction it can be approached — filling in missing definitions, spotting when "equally likely" genuinely applies, and moving back and forth between an event and its complement whenever only one of the two is given directly.
Filling In the Basic Facts
The first problem checks the vocabulary this whole chapter depends on, restating five relationships in blank form:
| Statement | Completed |
|---|---|
| P(E) + P(not E) = ___ | 1 |
| P(an impossible event) = ___ | 0 |
| P(a sure/certain event) = ___ | 1 |
| The sum of probabilities of all elementary events of an experiment is ___ | 1 |
| The probability of any event E lies between ___ and ___, inclusive | 0 and 1 |
Every one of these five facts reappears somewhere else in this exercise, dressed up as an actual scenario rather than a bare statement — the candy bag below leans on the impossible/certain pair, and the deck-of-cards problem later leans on the 0-to-1 boundary directly.
Is "Equally Likely" Always True?
Four everyday situations, each offering exactly two possible outcomes, test whether "two outcomes" automatically means "equally likely outcomes" — it doesn't, unless there's a genuine symmetry behind the two possibilities.
| Situation | Equally likely? | Why |
|---|---|---|
| A car starts, or it doesn't | No | Depends on fuel level, engine condition, and age of the car — no symmetry between the two outcomes |
| A basketball shot goes in, or it misses | No | Depends entirely on the player's skill — an expert shooter and a beginner don't share the same odds |
| A true-false answer is right, or wrong | Yes | Exactly two possible answers exist, with no reason to favour either one over the other |
| A newborn baby is a boy, or a girl | Yes | Two possible outcomes, each occurring with essentially the same underlying likelihood |
The pattern worth internalising here: "equally likely" is a claim about the mechanism producing the outcomes, not about how many outcomes there happen to be. Two outcomes influenced by skill, condition, or circumstance are not automatically 50-50 just because there are only two of them.
Complement From a Single Probability
When only one of P(E) or P(not E) is given, the complementary-event relationship supplies the other one instantly, with no further information needed at all.
If P(E) = 0.05, then P(not E) = 1 − 0.05 = 0.95A second problem applies exactly this same subtraction to a genuinely different-sounding scenario. Given that the probability of two students not sharing the same birthday is 0.992, the probability that they do share a birthday follows immediately as the complement, without needing to know anything about how the 0.992 figure was itself worked out:
P(same birthday) = 1 − P(not same birthday) = 1 − 0.992 = 0.008Notice that both problems use the identical one-line calculation despite looking unrelated on the surface — a decimal probability and a birthday-sharing scenario. Recognising when a question is secretly just "subtract from 1" is most of the work in this whole section, and it's worth noticing that neither problem needed the original 0.05 or 0.992 figure explained or re-derived — both were simply handed over as given data, exactly the way a real statistical study might report an already-computed probability without showing every step behind it.
A Bag of Lemon Candies
A bag contains only lemon-flavoured candies — no orange ones at all. Malini takes out one candy without looking. Because the bag physically contains nothing but lemon candies, both outcomes here sit at the extreme ends of the probability scale rather than somewhere in between:
P(an orange-flavoured candy) = 0 (impossible event — none exist in the bag)
P(a lemon-flavoured candy) = 1 (certain event — every candy in the bag is lemon)This is the impossible-event and sure-event facts from the very first table, now attached to a physical situation instead of stated abstractly. Whenever a sample space contains only one type of outcome, its probability is 1 and every excluded outcome's probability is 0 — no counting or ratio is even needed to see it. The two answers still add up correctly under the complementary-event rule, 0 + 1 = 1, even though this pairing looks nothing like the more typical fractional split seen in most other problems in this exercise.
A Shuffled Deck With the Hearts Removed
Rahim removes all thirteen hearts from a standard 52-card deck before shuffling the remaining 39 cards and drawing one at random. Removing a full suit changes the sample space size and, for hearts specifically, the favourable-outcome count too — every part of this problem has to work from the reduced 39-card deck, not the original 52.
| Event | Favourable outcomes | Probability |
|---|---|---|
| (i) An ace | 3 aces remain (the ace of hearts is gone) | 3/39 = 1/13 |
| (ii) A diamond | All 13 diamonds are untouched | 13/39 = 1/3 |
| (iii) Not a heart | All 39 remaining cards qualify | 39/39 = 1 (certain event) |
| (iv) The ace of hearts | Already removed from the deck | 0/39 = 0 (impossible event) |
Parts (iii) and (iv) are worth pausing on specifically, since they're the sure and impossible events from earlier in this exercise showing up a second time, now inside a 39-card deck rather than a candy bag — once every heart is physically gone from the deck, drawing "not a heart" is guaranteed and drawing "the ace of hearts" is guaranteed impossible, regardless of how the other 39 cards happen to be arranged.
Two More Direct Calculations
| Scenario | Favourable outcomes | Probability |
|---|---|---|
| Die rolled once — prime number | {2, 3, 5} out of {1,...,6} | 3/6 = 1/2 |
| Die rolled once — number lying between 2 and 6 | {3, 4, 5} out of {1,...,6} | 3/6 = 1/2 |
| Die rolled once — odd number | {1, 3, 5} out of {1,...,6} | 3/6 = 1/2 |
| One card drawn from a full 52-card deck — a red king | 2 red kings (hearts, diamonds) out of 52 | 2/52 = 1/26 |
All three parts of the die question happen to land on exactly the same probability, 1/2, despite counting three genuinely different subsets of outcomes — a coincidence specific to a standard six-sided die's particular numbers, not a general pattern that would hold for a die numbered differently or for a larger sample space.
The phrase "a number lying between 2 and 6" is worth reading carefully rather than skimming — taken literally, it excludes both endpoints, leaving only {3, 4, 5}, which is exactly how the table above treats it. A careless reading that quietly includes one or both endpoints would silently change n(E) from 3 to 4 or 5, and the final probability along with it. Whenever a problem's wording uses "between," "at least," or "at most," pinning down precisely which numbers are included before counting anything is worth the extra few seconds it takes.
Onward to a Longer Set of Problems
Every problem in this exercise turned on the single relationship P(E) + P(not E) = 1, applied to candies, cards, dice, and birthdays alike. Exercise 13.2 moves past that one relationship into a much longer run of classical-probability problems — coins, marbles, discs, spinners, and more — each one needing only the core formula P(E) = n(E)/n(S) from the chapter introduction, applied carefully to a new sample space each time.