Probability basics — sample space, events, P(E) = favourable - total: Question bank

This page is a misconception gym. Each line is a question ::: answer reveal. The answers carry the reasoning so you rebuild the idea, not just memorise a verdict.

The three pictures below are the only tools you need to unlock every trap on this page: a tree (why $H T \neq = T H$ ), a Venn diagram (why unions double-count), and a probability bar (why $E$ and $E^{'}$ fill the whole strip).

True or false — justify

The formula

P (E) = \frac{n ( E )}{n ( S )}

always works for any experiment.

False — it works only when every outcome in

S

is equally likely. For a bent coin or "number of heads in two tosses", equal-likelihood fails, so the ratio

\frac{n ( E )}{n ( S )}

gives the wrong number.

Two coins give outcomes

{0, 1, 2}

heads, so

P (1 head) = \frac{1}{3}

False — those three totals are not equally likely. The equally-likely space is

{H H, H T, T H, T T}

with

n (S) = 4

, so

P (1 head) = \frac{2}{4} = \frac{1}{2}

(the tree figure shows

H T

and

T H

as two separate branches).

An event is always a single outcome.

False — an event is any subset of the sample space, which may hold zero, one, or many outcomes (impossible, simple, or compound respectively).

P (E) = 0

then

E

can never possibly happen in principle.

In a finite equally-likely setup, yes —

P (E) = 0

means

n (E) = 0

, i.e.

E = \emptyset

. (In infinite spaces this weakens, but that is beyond this topic.)

P (E) = 0.5

means the event is guaranteed to occur in exactly half of any two trials.

False — probability describes the long-run fraction, not a guarantee over a tiny sample. Two flips can easily give two heads. See Law of Large Numbers.

For any event,

P (E) + P (E^{'}) = 1

True —

E

and its complement

E^{'}

don't overlap and together fill

S

(see the probability-bar figure), so their probabilities add to

P (S) = 1

Adding the probabilities of "even"

{2, 4, 6}

and "greater than 4"

{5, 6}

on a die gives

P (even or > 4)

False — the outcome

6

lives in both sets, so plain addition double-counts it. Use the general rule:

P (A \cup B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6}

, not

\frac{5}{6}

P (A) = 0.6

and

P (B) = 0.7

, then

P (A \cup B)

could be

1.3

False — probability can never exceed

1

(Axiom 1). The general rule subtracts

P (A \cap B)

, which must be at least

0.3

here, so the union stays at

1

or below.

Two events being mutually exclusive is the same as them being complementary.

False — Mutually Exclusive Events just means they can't both happen (

A \cap B = \emptyset

). Complementary events are mutually exclusive and together cover all of

S

; mutually exclusive events need not cover

S

The sample space for tossing two coins must always list

H H, H T, T H, T T

separately.

True if you want equally-likely outcomes.

H T

and

T H

are physically distinct (which coin showed heads), so collapsing them breaks equal likelihood.

Spot the error — each line hides one genuine mistake

"A bag has 5 red, 3 blue.

P (red) = \frac{5}{3}

Error: the denominator must be the total number of outcomes (

n (S) = 5 + 3 = 8

), not the count of the other colour. Correct:

\frac{5}{8}

P (sum = 7)

on two dice

= \frac{6}{12}

because there are 6 winning pairs and 12 faces."

Error: the sample space is ordered pairs, so

n (S) = 6 \times 6 = 36

, not

12

. Correct:

\frac{6}{36} = \frac{1}{6}

P (even OR > 4)

on a die

= \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

Error: even

{2, 4, 6}

and

> 4

{5, 6}

share the outcome

6

, so this ignores the overlap. General rule:

\frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}

"Face cards:

4

suits and

3

faces, so

n (E) = 4 + 3 = 7

out of

52

Error: you multiply, not add — for each of

4

suits there are

3

face cards, so

n (E) = 4 \times 3 = 12

, giving

\frac{12}{52} = \frac{3}{13}

P (E^{'}) = 1 - P (E)

only works when

E

is a simple event."

Error: the complement rule holds for any event, simple or compound — it depends only on

E

and

E^{'}

partitioning

S

, not on how many outcomes

E

has.

"To pick 2 balls from

{R, B, G}

, the sample space is just

{R B, R G, B G}

, so 3 outcomes always."

Error: this silently assumes unordered draws. If order matters (ball 1 vs ball 2),

n (S) = 6

R B, B R, R G, GR, B G, GB

. Fix the order-convention before counting. See Counting Principles.

Recall Which one is a "trick" (no real error)?

Where mutually-exclusive-and-add reasoning is honestly correct — a genuinely disjoint case. Is $P (even OR odd) = \frac{3}{6} + \frac{3}{6} = 1$ an error? ::: No — even and odd share no outcome ( $A \cap B = \emptyset$ ), so plain addition is exactly the disjoint special case and $1$ is correct. Only add directly when the sets are disjoint.

Why questions

Why must the sample space be settled before counting favourable outcomes?

Because

n (S)

is the denominator of every probability — a missed outcome quietly inflates every fraction and can push a "probability" above

1

Why does each equally-likely outcome get probability

\frac{1}{n ( S )}

All

n (S)

outcomes are equal (call each

p

) and must sum to

P (S) = 1

, so

n (S) \cdot p = 1

, giving

p = \frac{1}{n ( S )}

. Building block from Axioms of Probability.

Why is the complement rule often easier than direct counting?

For "at least one" problems, counting the single "none / all-failure" complement is one clean case, whereas the direct event splits into many overlapping cases. Since

E

and

E^{'}

fill the whole bar,

P (E) = 1 - P (E^{'})

Why can we add probabilities directly only when the parts are mutually exclusive?

Because the general rule

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

only loses its

- P (A \cap B)

term when

A \cap B = \emptyset

; otherwise shared outcomes get counted twice and the sum overstates the truth.

Why is

H T

counted as different from

T H

when tossing two coins, but

{R, B}

might equal

{B, R}

when drawing balls?

Coins are ordered (first vs second toss are distinguishable), so

H T \neq = T H

— two branches on the tree. Whether two draws are ordered is a modelling choice; state it, then count consistently.

Edge cases

What is

P (\emptyset)

and why?

P (\emptyset) = 0

— the impossible event contains no outcomes, so

n (E) = 0

and the ratio is

0

. It is the floor of the probability range.

What is

P (S)

and why?

P (S) = 1

— the sure event contains every outcome, so

n (E) = n (S)

and the ratio is

1

. Something must happen (Axiom 2).

If a die has

6

faces but you only care about "roll less than

7

", what is

P (E)

E = {1, 2, 3, 4, 5, 6} = S

, so

P (E) = 1

: it is a sure event dressed up as a condition.

Can two events be both mutually exclusive and not exhaustive? Give the idea.

Yes — on a die,

{1}

and

{2}

never overlap (mutually exclusive) yet leave

{3, 4, 5, 6}

uncovered (not exhaustive), so their probabilities do not sum to

1

What happens to the classical formula if outcomes are equally likely but there are infinitely many?

It breaks —

\frac{1}{n ( S )}

would be

\frac{1}{\infty}

, so each single outcome has probability

0

and you need a different (measure-based) tool. This topic assumes a finite equally-likely

S

If you draw a card, put it back, then draw again, does the second draw's sample space shrink?

No — with replacement the deck is restored, so

n (S) = 52

both times and draws are independent. Without replacement the second space shrinks to

51

; see Conditional Probability.

Recall One-line self-test

A "probability" you computed comes out to $1.4$ . What went wrong? ::: You either double-counted overlapping outcomes (forgot $- P (A \cap B)$ ) or used the wrong, too-small sample space — a real probability can never exceed $1$ .