Probability

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‘Equally likely’ is a claim, not an assumption

◆FRAME

Two candidates stand for one seat. It is tempting to call it a coin flip — two candidates, two outcomes, so a 50% chance each. That reasoning skips something real. A coin's two faces really are equally likely. An election's two candidates are not automatically equally likely to win. They are equally likely only if the underlying support genuinely is. That is exactly why pollsters measure support. They never assume it from the bare fact that two names sit on a ballot.

Probability's central formula rests on this same assumption, stated openly instead of smuggled in. The probability of an event is the ratio of outcomes favourable to that event, over the total number of equally likely outcomes in the sample space. Everything before "equally likely" is arithmetic — a count divided by a count. "Equally likely" itself is not arithmetic. It is a claim about the world. This chapter treats it as a claim to confirm, never one to assume.

A probability is a count divided by a count — but only once "equally likely" has been checked, not assumed.

The six faces are the same on both dice, so counting gives 1 over 6 for a six either way. Read the strips instead of the count — only the fair die splits the probability the way that count assumes, and the list above them is identical in both cases.

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Random experiments and the sample space

▲CONCEPT

Not every trial with an uncertain-looking result counts as a random experiment. Computing $17 \times 23$ by hand might feel uncertain if you have never done it before, but it is not random — the answer is fixed the moment the two numbers are chosen, and repeating the calculation always gives the identical result. A random experiment is different: even under identical conditions, repeating it can give a different outcome each time. What stays fixed beforehand is not the result, but the full list of results that could occur.

The formal names: a RANDOM EXPERIMENT is a trial whose outcome cannot be predicted with certainty, but whose set of POSSIBLE outcomes is known in advance. That set of all possible outcomes is the SAMPLE SPACE, written $S$. Each individual possible outcome inside it is a SAMPLE POINT.

The outcome stays unknown until the trial runs. The list of what it could be is known long before that.

Read the two rows by length. Both lists were written down before either trial ran, so knowing what could happen is not what separates them; what separates them is that the top list leaves the trial nothing to choose. That is why 17 times 23 is not a random experiment even for a reader who has not worked it out yet. No chip is tinted on either row, because no event has been named yet — that is the next section's job.

Worked example

Write the sample space for tossing two coins

first coin: $H$ or $T$
list what one coin alone can show
second coin: $H$ or $T$
list what the other coin alone can show
$S = \{H H, H T, T H, T T\}$
combine the two coins' outcomes — every pairing gets one sample point
$n(S) = 4$
the sample space's size is just how many sample points it lists

This 2 by 2 grid is the small case of a pattern the chapter leans on again and again: n independent coins give 2 to the n equally likely outcomes, and the same one-axis-per-coin idea is what later becomes the multiplication rule for counting a multi-stage experiment — one axis per stage, and the grid, or its higher-dimensional version, counts itself.

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Events: simple, compound, impossible, sure

▲CONCEPT

Roll a single die once. The full sample space is $S = \{1,2,3,4,5,6\}$ — six sample points, one for each face. An EVENT is any subset of this sample space: a well-defined description of which outcomes count.

A SIMPLE (or elementary) event contains exactly one sample point. "The die shows a $4$" is just $\{4\}$. A COMPOUND event contains more than one. "The die shows an even number" is $\{2,4,6\}$ — three sample points bundled under one description. Two special cases sit at the extremes. The IMPOSSIBLE event is $\emptyset$: it never occurs, like "the die shows a $7$." The SURE event is $S$ itself: it always occurs, since every outcome belongs to it.

Simple, compound, impossible, sure — four names for how large a subset of the sample space an event happens to be.

Two of these four labels are the endpoints every probability value sits between: impossible is defined to have probability 0, sure has probability 1, and every simple or compound event in between gets some fraction of the die's six faces.

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The algebra of events: probability borrows Chapter 1’s set language

▲CONCEPT

A scholarship notice that reads "open to students scoring above 90 percent OR holding a state-level sports certificate" names a UNION of two events, in plain notice-board phrasing. A hostel rule that applies specifically when a student is both a day-scholar AND repeatedly late names an INTERSECTION instead. Neither notice ever writes the word "set" — but both are already built on exactly this idea.

An event is a subset of the sample space, so Chapter 1’s set operations carry over directly. The COMPLEMENTARY event "not $A$" is $A'$. The event "$A$ or $B$" is $A \cup B$. The event "$A$ and $B$" is $A \cap B$. The event "$A$ but not $B$" is $A - B$. Probability's event vocabulary is not new vocabulary. It is Chapter 1’s set language, applied to outcomes instead of arbitrary objects.

Every "or," "and," "not," and "but not" in an event description already names a set operation.

Every rule the rest of this chapter builds on events — mutually exclusive, exhaustive, the addition rule — is a statement about one of these three operations applied to a specific pair of events. Once union, intersection and complement can be read straight off a chip diagram, none of those rules needs a diagram of its own; each one reuses this.

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Mutually exclusive and exhaustive events

▲CONCEPT

A single student's attendance record for one day reads PRESENT or ABSENT, never both at once — the everyday shape of what MUTUALLY EXCLUSIVE means for two events.

Two events $A$ and $B$ are MUTUALLY EXCLUSIVE when $A \cap B = \emptyset$. They share no outcome, so they can never both occur on the same trial. On a single die roll, "the result is even" and "the result is odd" are mutually exclusive. No outcome belongs to both.

Two events are mutually exclusive exactly when their intersection is empty — nothing satisfies both descriptions at once.

This is a check, not a computation: list each event's outcomes and look for one that appears in both. The top row has none to find; the bottom row has exactly one, at 6. That single shared outcome is enough on its own to break mutual exclusivity, however many outcomes the two events otherwise fail to share.

▲CONCEPT

A city's postal-code zones illustrate the same idea. Every address falls inside at least one zone, and the full set of zones covers the entire city, leaving no address unassigned.

A collection of events $A_1, A_2, \ldots, A_n$ is exhaustive when their union is the whole sample space: $A_1 \cup A_2 \cup \ldots \cup A_n = S$. Together they cover every possible outcome, so at least one of them must occur on every trial.

Mutually exclusive events cannot overlap. Exhaustive events cannot leave a gap. A collection can do either, both, or neither.

Mutually exclusive and exhaustive together are the two conditions a set of events needs before its probabilities are guaranteed to add up to exactly 1 — drop exhaustive, as the bottom row does, and the probabilities of the pieces still present can add to less than 1, with the gap sitting in whatever was left uncovered. Both properties have to hold at once; the complement rule elsewhere in this chapter is the special case with only two pieces, where both are guaranteed automatically.

Worked example

Check whether two die-roll events are mutually exclusive, exhaustive, or both

$A = \{1,2,3\}$, $B = \{4,5,6\}$
the two events being checked, on a single die roll
$A \cap B = \emptyset$
no outcome belongs to both sets — the events are mutually exclusive
$A \cup B = \{1,2,3,4,5,6\} = S$
together the two events cover every outcome — the events are exhaustive
both properties hold at once
true for this pair, but not guaranteed for every event pair

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Axiomatic probability: turning a count into a likelihood

▲CONCEPT

Suppose a basket holds 12 fruits, of which 5 are apples. Let $E$ be the event "the fruit is an apple": $n(E) = 5$, and $n(S) = 12$. So $P(E) = 5/12$. Every number here comes straight from counting, once each fruit is equally likely to be the one picked.

For a sample space $S$ whose outcomes are confirmed equally likely, the PROBABILITY of an event $E$ is

$P(E) = n(E)/n(S),$

where $n(E)$ counts outcomes favorable to $E$ and $n(S)$ counts every outcome in $S$. Probability always satisfies $0 \leq P(E) \leq 1$, with $P(S) = 1$ (the sure event) and $P(\emptyset) = 0$ (the impossible event).

A probability is a favourable count divided by a total count — but only once the sample space's outcomes are confirmed equally likely.

Every rule later in this chapter is a way of combining counts built exactly this way, without starting the count over: the complement rule and the addition rule both lean on ratios of marked outcomes over the same fixed six, never a fresh count from nothing.

Worked example

Find the probability of rolling a number greater than 4

favourable outcomes: $\{5, 6\}$
list every outcome the event actually includes
$n(E) = 2$
count the favourable outcomes
$n(S) = 6$
count every outcome in the sample space
$P(E) = 2/6 = 1/3$
favourable outcomes over total outcomes, then simplify

Read this against the count-over-count figure a few pages back. There, n(S) stayed at 6 and the marked count moved, so the answer moved with it. Here the count stays at 2 and the marked faces move instead, and the answer does not budge. Between them the two figures say which of the two numbers in P(E) = n(E)/n(S) is doing the work in any given question.

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The probability of the complement

▲CONCEPT

Continue the fruit-basket example. Picking an apple has probability $5/12$. What is the probability of NOT picking an apple? No separate count of the other fruits is required — subtract from $1$ instead: $1 - 5/12 = 7/12$. Apple and not-apple between them account for every fruit in the basket, with nothing left over and nothing double-counted.

$A$ and its complement $A'$ are always mutually exclusive of each other, $A \cap A' = \emptyset$, and exhaustive of each other, $A \cup A' = S$. So $P(A) + P(A') = P(S) = 1$, which rearranges to

$P(A') = 1 - P(A) \cdot$

Finding "the probability that $A$ does not happen" never needs its own separate count. It is always one minus $P(A)$.

_$A$ and its complement split the sample space exactly, with nothing left over and nothing shared — which is why their probabilities always add to one._

This is the shortcut worth reaching for whenever a question asks for ‘at least one’ of something across several trials: counting every way to get at least one head in ten coin tosses directly is painful, but its complement, no heads at all, is a single outcome to count. The complement rule turns a hard sum of many cases into one probability subtracted from 1, and ‘at least one’ is the exam phrase that should make you reach for it.

Worked example

Find the probability of NOT rolling a 6

$S = \{1,2,3,4,5,6\}$, $n(S) = 6$
a fair die's six equally likely outcomes
getting a $6$: $P = 1/6$
one favourable outcome out of six
not getting a $6$: $P = 1 - 1/6$
the complement rule — subtract from one instead of counting the other five outcomes directly
$1 - 1/6 = 5/6$
the probability of not getting a 6

The rule doing the work here is P(not A) = 1 − P(A), which holds no matter how big the population gets. That is the entire reason it is worth having as a rule rather than a counting habit: scale the bag up a hundredfold and the subtraction still takes one line, while a direct count would not.

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The additive rule: correcting for a double-counted overlap

▲CONCEPT

Estimating the probability that a randomly chosen employee is either in Engineering or has more than five years' experience has to subtract the overlap — the employees who are both — exactly once. Skip that subtraction, and every employee who is both gets counted twice.

For any two events $A$ and $B$:

$P(A \cup B) = P(A) + P(B) - P(A \cap B) \cdot$

This is the probability version of Chapter 1’s inclusion-exclusion counting formula, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$, with every count divided through by $n(S)$. Adding $P(A)$ and $P(B)$ counts every outcome in $A \cap B$ twice. $P(A \cap B)$ is subtracted once to correct it.

Add the two probabilities, then subtract the overlap once — the same correction Chapter 1 already made for counts, now made for probabilities.

This is the general rule the earlier mutually-exclusive figure was a special case of: when two events cannot share an outcome, P of A intersect B is already zero, so the subtraction vanishes on its own and just adding the two probabilities works. Checking for a shared outcome first, the way this deck was checked, is the actual first step of every union problem; the subtraction only ever corrects for what that check finds.

Worked example

Find the probability of drawing a King or a Heart from a 52-card deck

$A$ = King, $n(A) = 4$
the first event and its outcome count
$B$ = Heart, $n(B) = 13$
the second event and its outcome count
$A \cap B$ = King of Hearts, $n(A \cap B) = 1$
the one card counted in both events — this is the overlap the rule corrects for
$P(A \cup B) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13$
add the two probabilities, subtract the overlap once, then simplify

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The additive rule when there is nothing to correct

▲CONCEPT

When $A$ and $B$ are MUTUALLY EXCLUSIVE, $A \cap B = \emptyset$. So $P(A \cap B) = 0$, and the additive rule simplifies to plain addition:

$P(A \cup B) = P(A) + P(B) \cdot$

This is not a different rule from the last section's. It is the same general additive rule, with the correction term dropped because there is nothing left to correct.

Mutually exclusive events share no overlap to subtract — the general rule and the simple rule agree exactly, because the correction term is already zero.

Worked example

Find the probability the result is 1 or 2 on a single die roll

$A = \{1\}$, $B = \{2\}$
the two events being combined
$A \cap B = \emptyset$
no outcome is both — the events are mutually exclusive
$P(A \cup B) = P(A) + P(B)$
no overlap to subtract, so the additive rule reduces to plain addition
$1/6 + 1/6 = 2/6 = 1/3$
add the two probabilities directly and simplify

What to look for here is a colour that never appears. Step 2 of the worked example is the line A intersect B equals the empty set, which on the page is algebra to be believed; on these rows it is the plain absence of any chip carrying the both colour. Do the try-it row before reading its answer: it names two different faces and lands on the same third, because what licenses the plain sum is the absence, not which two faces were picked.

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Every idea, one question

■RECAP

Every idea in this chapter answers one question: how likely, exactly?

An event is a subset of the sample space. Chapter 1’s set operations name every combination of events without new vocabulary. $P(E) = n(E)/n(S)$ turns a count into a likelihood, once the outcomes are confirmed equally likely rather than assumed to be. The complement rule and the additive rule both let a hard probability be built from easier ones, instead of counted from scratch. The additive rule's mutually-exclusive special case is the ordinary case with nothing left to correct for.

In practice, start by naming what a problem actually hands you. A plain count of favourable versus total outcomes, once equally-likely is confirmed? Use $P(E) = n(E)/n(S)$ directly. Asked for the chance something does NOT happen? Subtract from one instead of recounting. Asked for an "or" across two events that might overlap? Add the two probabilities and subtract the overlap once. Told the two events share no outcome at all? Skip the subtraction — there is nothing left to correct.

Every rule in this chapter exists because "equally likely" has to be earned before it can be divided by.

Read down the rows, not across them. The chapter's four rules arrive in the prose as four things to remember; here they are four markings of one unchanging sample space, so the only question a problem ever poses is which chips to tint and what to do with the counts. Rows three and four are the pair worth comparing: they are the same rule, and the fourth row's zero is not a shortcut but the correction term arriving as nothing.

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Two traps this chapter sets

✕MISCONCEPTION

THE TRAP. $P(A \cup B) = P(A) + P(B)$, always — just add the two probabilities.

THE REALITY. That only works when $A$ and $B$ are mutually exclusive. Whenever they overlap, every outcome in $A \cap B$ was already counted once inside $P(A)$, and again inside $P(B)$. $P(A \cap B)$ must be subtracted once to correct the double-count — the same boundary error as Chapter 1’s counting-formula misconception, now carried into probability.

_Skip the subtraction whenever two events overlap, and the computed "probability" is too high by exactly $P(A \cap B)$ — not a small rounding error, a real double-count._

The same two events, added twice over

Weaker. In a group of $40$ people, $18$ wear glasses and $12$ cycle to work. Asked for $P(\text{glasses or cycles})$, the quick move is to add the two probabilities as they stand: $18/40 + 12/40 = 30/40 = 3/4$. Nothing about that line looks wrong. It is the same addition that works for a die's even numbers and its multiples of $5$, and it takes both given numbers and uses them. But it has not asked whether any of these $40$ people were counted on both lists.

Stronger. Five of them were: $5$ people both wear glasses and cycle. Those five sit inside the $18$ and inside the $12$, so the plain sum counts each of them twice, and $P(A \cap B) = 5/40$ has to come off once: $18/40 + 12/40 - 5/40 = 25/40 = 5/8$. Count the people instead of the events and the same $25$ appears --- $13$ who only wear glasses, $5$ who do both, $7$ who only cycle. The subtraction did not remove those five from the answer; it removed the second count of them.

The chapter's mutually-exclusive worked example got away with plain addition because its two events shared no outcome to double-count. This one does, which is exactly the additive rule's job: subtract once for whatever the plain sum counted twice — never zero times, and never twice.

✕MISCONCEPTION

THE TRAP. A sample space with $n$ listed outcomes automatically gives each outcome probability $1/n$.

THE REALITY. $P(E) = n(E)/n(S)$ only holds when the outcomes are CONFIRMED equally likely, by the experiment's own symmetry — a fair coin, a fair die. Listing outcomes never by itself makes them equally likely. A biased die has a well-defined sample space too. So does a race between runners of different speeds. Neither one's outcomes are equally likely, and dividing by $n$ there gives a wrong answer.

A sample space lists what could happen. It says nothing about how often each listed outcome actually does — that has to be checked separately, every time.

Contrast this with the coin and dice figures earlier in the chapter: those outcomes are equally likely because of physical symmetry, a coin has no reason to favour one face and a die no reason to favour one number, and that reasoning holds before a single trial is run. Runners have no such symmetry to appeal to. The only way to find their real chances is to time enough races and count, which is why this pair of strips needed two different sources instead of one shared list.

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Practice set

Exercise 14.1 — Sample spaces and events

practice A coin is tossed three times. Describe the event "exactly two heads" as a subset of the sample space, and state $n(E)$.
practice A die is rolled once. Which of the following is the SURE event?
1. $\{1, 2, 3\}$
2. $\{7\}$
3. $\{1, 2, 3, 4, 5, 6\}$
4. $\emptyset$
practice A die is rolled once. Let $A$ = "the number is even" and $B$ = "the number is a multiple of $3$". Find $A \cap B$ and $A \cup B$.
practice Two events $A$ and $B$ satisfy $A \cap B = \emptyset$. Which term describes this relationship?
1. mutually exclusive
2. exhaustive
3. complementary
4. equally likely
practice A card is drawn from a standard deck. Let $A$ = "the card is a Spade" and $B$ = "the card is a Heart". Are $A$ and $B$ mutually exclusive? Are they exhaustive?
practice Two dice are thrown together. Write the event "the sum is at least $11$" as a subset of the sample space, and state $n(E)$.

Answers

$E = \{H H T, H T H, T H H\}$, $n(E) = 3$
$\{1, 2, 3, 4, 5, 6\}$
$A \cap B = \{6\}$, $A \cup B = \{2, 3, 4, 6\}$
mutually exclusive
Mutually exclusive; not exhaustive.
$E = \{(5,6), (6,5), (6,6)\}$, $n(E) = 3$

Problem 5 asks two questions about the same pair of events, and the answers differ. Run a finger down any column: no card is tinted twice. Then look at what is left plain — half the deck belongs to neither event.

Sorted by what they ask rather than by number, the six turn out to be three questions asked twice. That matters when you get one wrong: the fix is one of three things to re-read, not six problems to redo. Note what is missing from the whole table — no probability is asked for anywhere in Exercise 14.1. Every answer here is a set, a count or a name.

Exercise 14.2 — Axiomatic probability

practice A bag contains $5$ red balls and $3$ blue balls. One ball is drawn at random. Find the probability that it is red.
practice A card is drawn from a well-shuffled deck of $52$ cards. What is the probability that it is a King?
1. $1/13$
2. $1/52$
3. $4/13$
4. $1/4$

Answers

$5/8$
$1/13$

Both problems are the same three lines of work — mark, count, divide — on sample spaces six times apart in size. The one thing the size changes is whether the answer arrives finished. Read the ticks under the deck rather than the fraction: 4 out of 52 cards and 1 out of 13 ranks are the same marking, counted at two different grains. That is what reducing a probability means here.

Miscellaneous — Complement and additive rules

practice A die is rolled once. Find the probability of NOT getting a multiple of $3$.
practice In a class of $50$ students, $20$ play cricket, $15$ play football, and $8$ play both. Find the probability that a randomly selected student plays cricket or football.
practice For two mutually exclusive events $A$ and $B$ with $P(A) = 0.3$ and $P(B) = 0.4$, what is $P(A \cup B)$?
1. $0.12$
2. $0.7$
3. $0.1$
4. $1$

Answers

$1 - 2/6 = 2/3$
$20/50 + 15/50 - 8/50 = 27/50$
$0.7$

This is the miscellaneous exercise's problem 1, done both ways at once. Cover the read-outs and count each row yourself first: the rows tint different chips and finish at the same number, which is the whole content of the complement rule. Six faces are small enough that neither route costs anything, and that is the point of meeting the rule here rather than on the 52-card problems where it starts to matter.

Count the tinted chips on the bottom row before reading its fraction. The problem gives 20 and 15 in one breath and the 8 in the next, which is why the wrong answer 35 over 50 feels reasonable while it is being written. The bottom row cannot say 35 — there are only fifty people in it and twenty-three of them are plain — so the subtraction stops being a rule to remember and becomes the eight students who were already on the row.

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Chapter-end problems

Chapter-end problems — graded set

board-easy Two coins are tossed together. Write the sample space and find P(exactly one head).
board-easy A fair die is thrown once. Find P(getting a number greater than 4).
board-easy A card is drawn from a well-shuffled deck of 52 cards. Find P(the card is a king).
board-easy A bag has 5 red and 3 blue balls. One ball is drawn at random. Find P(it is red).
board-easy Two dice are thrown together. Find P(the sum of the numbers is 7).
board-easy A coin is tossed 3 times. Find P(exactly two heads).
board-easy A number is chosen at random from 1 to 20. Find P(the number is a multiple of 3).
1. $1/3$
2. $3/10$
3. $7/20$
4. $3/20$
board-easy A die is thrown once. Find P(not getting a 6), using the complement rule.
board-easy Events A and B are mutually exclusive with P(A)=0.3 and P(B)=0.4. Find P(A or B).
board-easy From a deck of 52 cards, one card is drawn. Find P(the card is not a heart), using the complement rule.
board-standard Two dice are thrown together. Find P(the sum of the numbers is a multiple of 3).
board-standard From a well-shuffled deck of 52 cards, one card is drawn. Find P(the card is a face card or a heart).
1. $25/52$
2. $11/26$
3. $19/52$
4. $1/2$
board-standard Three coins are tossed together. Find P(at least two heads).
board-standard In a class, P(likes Maths) = 0.5, P(likes Science) = 0.4, and P(likes both) = 0.2. Find P(likes Maths or Science).
board-standard A committee of 3 is chosen at random from 5 men and 4 women. Find P(all three are men).
board-standard Two dice are thrown. Find P(a doublet), that is, both dice show the same number.
board-standard A number is chosen at random from 1 to 100. Find P(the number is divisible by 5 or 7).
board-standard A card is drawn from a deck of 52. Find P(the card is a spade or an ace).
1. $17/52$
2. $1/4$
3. $4/13$
4. $5/13$
board-standard For events A and B, P(A) = 0.6, P(A union B) = 0.8 and P(A inter B) = 0.3. Find P(B).
1. $0.6$
2. $0.7$
3. $0.4$
4. $0.5$
board-standard A die is thrown twice. Find P(the sum of the two numbers is at least 10).
board-standard A committee of 4 is chosen at random from 6 boys and 4 girls. Find P(exactly two girls are selected).
board-standard Events A and B satisfy P(A) = 0.42, P(B) = 0.48 and P(A inter B) = 0.16. Find P(A').
JEE For events A and B, P(A)=0.4, P(B)=0.3 and P(A union B)=0.58. Find P(A inter B) and state whether A and B are mutually exclusive.
JEE Two dice are thrown. Let A be the event that the sum is even and B the event that the sum exceeds 8. Find P(A union B).
JEE A box has 6 red, 4 green and 5 blue balls. Two balls are drawn at random without replacement. Find P(both balls are the same colour).
JEE Events A and B are mutually exclusive. P(A)=x and P(B)=2x. If P(A or B)=0.6, find x.
1. $0.1$
2. $0.2$
3. $0.3$
4. $0.15$
JEE Four cards are drawn at random from a well-shuffled deck of 52. Find P(all four are aces).
JEE A 5-digit code is formed from the digits 1 to 9, without repetition. Find P(the code is divisible by 5).
JEE For events A and B, P(A)=0.5 and P(B)=0.6. Using $0 \leq P(A \cup B) \leq 1$, find the range of possible values of P(A inter B).
JEE A committee of 5 is formed from 7 men and 5 women. Find P(the committee has at least 3 women).

Answers

$1/2$
$1/3$
$1/13$
$5/8$
$1/6$
$3/8$
$3/10$
$5/6$
$0.7$
$3/4$
$1/3$
$11/26$
$1/2$
$0.7$
$5/42$
$1/6$
$8/25$
$4/13$
$0.5$
$1/6$
$3/7$
$0.58$
$P(A \cap B) = 0.12$, not mutually exclusive
$2/3$
$31/105$
$0.2$
$1/270725$
$1/9$
$0.1 \leq P(A \cap B) \leq 0.5$
$41/132$

Both chapter-end problems begin the same way — write the sample space — and both are lost the same way, by listing outcomes as descriptions rather than as results. Look at HT and TH on the top row: in English they are one thing, on the row they are two chips, and a reader who merges them divides by 3 instead of 4. The bottom row is the same trap one size up, and the two row lengths are the 2-to-the-n rule the margin states.

Problem 29 hands you no outcomes to count, so slide instead. Neither bar can change length: push them apart until the strip is full, then together until A is swallowed, and read the shared cell at each stop.

The two rows are deliberately the same width, so what changes between them is the grain and nothing else. On six faces the plain chips and the tinted ones are equally quick to count, which is why problem 8 can be done without the rule at all. On fifty-two the plain block is one glance and the tinted block is not, and that difference — not any new mathematics — is what the complement rule buys.

Problem 17 asks this on a hundred numbers, where no reader can check the overlap by looking; the bottom row runs the identical move on twenty, where the overlap is one chip and can be checked. Find 15 before reading the fraction — it is the only number on both lists, and adding the two lists without it gives ten marked chips where the row shows nine. Over 1 to 100 the same shared numbers are the multiples of 35, and there are two of them.

These are the two chapter-end problems that hand you probabilities instead of outcomes, so there is nothing to count and the widths do the counting instead. Put a finger on each strip's dashed line: they land in the same place, which is why both answers are 0.7. Then read what sits to the left of each — the lower strip's 0.5 and 0.4 overlap on the middle cell, and the 0.2 that gets subtracted is that cell, laid down once rather than twice.

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JEE-application problems

JEE-application problems — from the item bank

Two dice are thrown together. Find P(the sum of the two numbers is 8).
1. 4/36
2. 5/36
3. 6/36
4. 1/36
Check your answer
1. 4/36 — 4/36 misses one of the five pairs that sum to eight.
2. ✓ 5/36 — (B) The pairs (2,6),(3,5),(4,4),(5,3),(6,2) all sum to 8 — five pairs out of thirty-six.
3. 6/36 — 6/36 counts six pairs, one too many — only five pairs sum to eight.
4. 1/36 — 1/36 counts only one ordered pair, forgetting the other four.
A card is drawn from a deck of 52. Find P(it is a queen or a spade).
1. 1/4
2. 4/13
3. 17/52
4. 12/52
Check your answer
1. 1/4 — 13/52 = 1/4 counts spades alone, ignoring the three queens that are not spades.
2. ✓ 4/13 — (B) 4 queens + 13 spades minus 1 shared card = 16 cards, so P = 16/52 = 4/13.
3. 17/52 — 17/52 adds 4 queens and 13 spades without correcting for the queen of spades.
4. 12/52 — 12/52 removes the queen of spades from both counts, losing a valid card.
A bag has 4 red and 6 black balls. Two balls are drawn at random without replacement. Find P(both balls are red).
1. 4/25
2. 3/50
3. 2/15
4. 4/15
Check your answer
1. 4/25 — 4/25 treats the draws as with replacement — but the second ball comes from a smaller bag.
2. 3/50 — 3/50 miscounts the total pairs; C(10,2) is 45, not 50.
3. ✓ 2/15 — (C) C(4,2)/C(10,2) = 6/45 = 2/15.
4. 4/15 — 4/15 uses an ordered count for the red balls, not the unordered pair count.
Events A and B satisfy P(A)=0.3, P(B)=0.5 and P(A and B)=0.1. Find P(A or B).
1. 0.6
2. 0.7
3. 0.9
4. 0.8
Check your answer
1. 0.6 — 0.6 subtracts the overlap twice instead of once.
2. ✓ 0.7 — (B) 0.3 + 0.5 minus 0.1 = 0.7 — add the two probabilities, subtract the overlap once.
3. 0.9 — 0.9 adds all three numbers instead of correcting for the overlap.
4. 0.8 — 0.8 adds P(A) and P(B) without subtracting the 0.1 overlap.
A committee of 3 is chosen at random from 4 men and 3 women. Find P(at least one woman).
1. 18/35
2. 12/35
3. 4/35
4. 31/35
Check your answer
1. 18/35 — 18/35 is P(exactly one woman) — “at least one” also counts two and three women.
2. 12/35 — 12/35 is P(exactly two women) — only part of “at least one”.
3. 4/35 — 4/35 is P(no woman) — the complement of what was asked.
4. ✓ 31/35 — (D) P(no woman) = C(4,3)/C(7,3) = 4/35, so P(at least one) = 1 minus 4/35 = 31/35.
A die is thrown twice. Find P(the two numbers are different).
1. 5/6
2. 5/36
3. 1/6
4. 25/36
Check your answer
1. ✓ 5/6 — (A) P(same) = 6/36 = 1/6, so P(different) = 1 minus 1/6 = 5/6.
2. 5/36 — 5/36 never subtracts the doublet count from one.
3. 1/6 — 1/6 is P(same number) — the complement of what was asked.
4. 25/36 — 25/36 treats the two dice independently instead of counting outcomes directly.
Two coins are tossed together. Find P(at least one tail).
1. 1/2
2. 3/4
3. 3/8
4. 1/4
Check your answer
1. 1/2 — 1/2 does not match 1 minus 1/4; recompute the complement.
2. ✓ 3/4 — (B) P(no tail) = P(HH) = 1/4, so P(at least one tail) = 1 minus 1/4 = 3/4.
3. 3/8 — 3/8 undercounts; three of the four outcomes have at least one tail.
4. 1/4 — 1/4 is P(no tail), namely HH — the complement of what was asked.
From a group of 3 boys and 2 girls, two people are selected at random. Find P(both are boys).
1. 1/10
2. 3/5
3. 1/5
4. 3/10
Check your answer
1. 1/10 — 1/10 undercounts; there are 3 boy-only pairs, not 1.
2. 3/5 — 3/5 is P(at least one girl selected), not P(both boys).
3. 1/5 — 1/5 does not match 3/10; recheck the simplification.
4. ✓ 3/10 — (D) C(3,2)/C(5,2) = 3/10.
Events A and B are mutually exclusive with P(A)=0.25 and P(A or B)=0.6. Find P(B).
1. 0.6
2. 0.85
3. 0.35
4. 0.15
Check your answer
1. 0.6 — 0.6 is P(A or B) itself, not P(B).
2. 0.85 — 0.85 adds 0.25 and 0.6 instead of subtracting to isolate P(B).
3. ✓ 0.35 — (C) P(A or B) = P(A) + P(B) for mutually exclusive events, so P(B) = 0.6 minus 0.25 = 0.35.
4. 0.15 — 0.15 does not match 0.6 minus 0.25; recheck the subtraction.
A number is chosen at random from 1 to 30. Find P(the number is a multiple of 4 or 6).
1. 4/15
2. 1/6
3. 2/5
4. 1/3
Check your answer
1. 4/15 — 4/15 does not match 10/30; recheck the simplification.
2. 1/6 — 1/6 undercounts; there are ten favourable numbers, not five.
3. 2/5 — 12/30 = 2/5 adds 7 and 5 without removing the 2 multiples of 12 counted twice.
4. ✓ 1/3 — (D) 7 multiples of 4 + 5 multiples of 6 minus 2 multiples of 12 = 10, so P = 10/30 = 1/3.
Three coins are tossed together. Find P(no heads at all).
1. 1/4
2. 3/8
3. 1/8
4. 7/8
Check your answer
1. 1/4 — 1/4 does not match 1/8; recheck the outcome count.
2. 3/8 — 3/8 is P(exactly one head), a different event.
3. ✓ 1/8 — (C) Only TTT has no heads, so P = 1/8.
4. 7/8 — 7/8 is P(at least one head) — the complement of what was asked.
A card is drawn from a deck of 52. Find P(it is neither a king nor a queen).
1. 11/13
2. 12/13
3. 1/13
4. 2/13
Check your answer
1. ✓ 11/13 — (A) P(king or queen) = 4/52 + 4/52 = 8/52 = 2/13, so P(neither) = 1 minus 2/13 = 11/13.
2. 12/13 — 12/13 uses only 4 cards in the complement, not the full 8 kings and queens.
3. 1/13 — 1/13 does not match 1 minus 2/13; recheck the subtraction.
4. 2/13 — 2/13 is P(king or queen), not its complement.
For mutually exclusive events A, B with P(A)=0.5 and P(B)=0.4, find P(A' and B').
1. 0.9
2. 0.19
3. 0.5
4. 0.1
Check your answer
1. 0.9 — 0.9 is P(A or B) itself, not the complement asked for.
2. 0.19 — 0.19 does not match 1 minus 0.9; recheck the subtraction.
3. 0.5 — 0.5 is P(A) alone, unrelated to the intersection of the two complements.
4. ✓ 0.1 — (D) P(A or B) = 0.5+0.4 = 0.9, and A' and B' is the complement of A or B, so P = 1 minus 0.9 = 0.1.
Two dice are thrown. Find P(the product of the two numbers is 12).
1. 1/9
2. 1/6
3. 1/18
4. 1/12
Check your answer
1. ✓ 1/9 — (A) (2,6),(3,4),(4,3),(6,2) all give product 12 — four pairs out of thirty-six, so P = 1/9.
2. 1/6 — 6/36 = 1/6 includes pairs whose product is not actually 12.
3. 1/18 — 2/36 = 1/18 misses two of the four pairs whose product is 12.
4. 1/12 — 1/12 does not match 4/36; recheck the simplification.
A committee of 4 is formed from 5 engineers and 3 doctors. Find P(the committee has exactly 2 doctors).
1. 4/7
2. 3/7
3. 1/7
4. 2/7
Check your answer
1. 4/7 — 4/7 overcounts by including committees with more than two doctors.
2. ✓ 3/7 — (B) C(3,2) times C(5,2) = 3 times 10 = 30, so P = 30/70 = 3/7.
3. 1/7 — 10/70 = 1/7 undercounts; there are 30 committees with exactly two doctors, not 10.
4. 2/7 — 2/7 does not match 30/70; recheck the simplification.
If P(A)=0.6 and P(B)=0.5, find the maximum possible value of P(A and B).
1. 0.5
2. 0.1
3. 1.0
4. 0.6
Check your answer
1. ✓ 0.5 — (A) A and B is always a subset of both A and B, so P(A and B) can be at most the smaller of the two, 0.5.
2. 0.1 — 0.1 is the minimum possible value of P(A and B), not the maximum.
3. 1.0 — 1.0 ignores that A and B is a subset of both A and B, so its probability is capped.
4. 0.6 — 0.6 is P(A) itself, not the bound on the smaller intersection.
If P(A)=0.6 and P(B)=0.5, find the minimum possible value of P(A and B).
1. 0.5
2. 0
3. 0.6
4. 0.1
Check your answer
1. 0.5 — 0.5 is the maximum possible value of P(A and B), not the minimum.
2. 0 — 0 would need P(A or B) to reach 1.1, but a probability never exceeds 1.
3. 0.6 — 0.6 is P(A) itself, not the bound on the intersection.
4. ✓ 0.1 — (D) P(A or B) = 1.1 minus P(A and B) can be at most 1, so P(A and B) is at least 0.1.

The counting is the difficulty, not the probability: pick the doctors, pick the engineers, multiply.

A wrong answer in this row is worth diagnosing before moving on: if the mistake was in the fraction itself, the fix is in this chapter; if the mistake was in what went into the numerator or denominator, this chapter cannot fix it and the count itself needs revisiting.

Whenever a question's sample space is a list of totals or combined outcomes rather than individual choices, check first whether that list was built from equally likely elements before assigning 1 over its length to any one entry. Dice values are equally likely; dice sums are not, and neither are card totals or drawn-card combinations in general.

JEE-application problems — from the item bank (continued)

A fair die is rolled once. Find P(the number shown is prime or even).
1. 1
2. 5/6
3. 2/3
4. 1/3
Check your answer
1. 1 — 1 adds 3 primes and 3 evens without removing the shared outcome, 2.
2. ✓ 5/6 — (B) Primes 2, 3, 5; evens 2, 4, 6; together 2, 3, 4, 5, 6 — five faces out of six.
3. 2/3 — 2/3 undercounts; five of the six faces are prime or even.
4. 1/3 — 1/3 is P(the number is prime) alone, ignoring the even faces.
A box has 3 defective and 7 non-defective bulbs. Two bulbs are drawn at random without replacement. Find P(at least one is defective).
1. 1/15
2. 8/15
3. 1/5
4. 7/15
Check your answer
1. 1/15 — 1/15 undercounts; there are 21 non-defective pairs, not 3.
2. ✓ 8/15 — (B) P(none defective) = C(7,2)/C(10,2) = 21/45 = 7/15, so P(at least one) = 1 minus 7/15 = 8/15.
3. 1/5 — 1/5 does not match 1 minus 7/15; recheck the subtraction.
4. 7/15 — 21/45 = 7/15 is P(no defective) — the complement of what was asked.
Events A and B satisfy P(A)=0.7, P(B)=0.6 and P(A or B)=1. Find P(A and B).
1. 0.3
2. 1.3
3. 0.1
4. 0.7
Check your answer
1. ✓ 0.3 — (A) P(A and B) = P(A)+P(B)-P(A or B) = 0.7+0.6-1 = 0.3.
2. 1.3 — 1.3 adds P(A) and P(B) but never subtracts P(A or B).
3. 0.1 — 0.1 does not match 1.3 minus 1; recheck the subtraction.
4. 0.7 — 0.7 is P(A) itself, not the intersection.