AFCAT Numerical Ability · Probability

Coin Tosses: Counting Heads, At Least One, Order and Given That

n fair coins, or one coin tossed n times, give 2ⁿ equally likely outcomes; count the favourable ones by listing, by choosing which tosses are heads, or by taking away the ones you do not want.

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Why this matters

The page asks for exactly, at least, at most and patterns of heads among up to five coins, and about a fifth of it is HARD. The HARD ones are about order: a first head on a given toss, heads side by side, or a condition you are given. Marks go when a listed sample space misses an outcome, or when 'at least two' is counted as 'exactly two'.

Concept 1 of 5: Listing the outcomes for up to three coins

Each coin has 2 faces, so 2 coins give 4 outcomes and 3 coins give 8. List them in a fixed order so none is missed: for 3 coins, HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

Definition

n coins tossed together, or one coin tossed n times, give $2^n$ equally likely outcomes: 2, 4, 8, 16, 32 for n = 1 to 5.
HT and TH are different outcomes. Treating 'one head and one tail' as a single outcome gives a wrong total of 3 for two coins.
Build the list by branching: write every outcome that starts with H, then every one that starts with T.
Read the favourable outcomes off the list, then divide by $2^n$ .

Outcomes for n coins

n(S) = 2^n, \qquad P(E) = \dfrac{n(E)}{2^n}

Worked example

Three coins are tossed. Find the probability of exactly one head.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Probability · Coin Tosses

What is the probability of getting an odd number of heads when three fair coins are tossed?

HT and TH are two outcomes

Two coins have four outcomes, not three. 'One head and one tail' happens in 2 of the 4, so its probability is

\tfrac12

, not

\tfrac13

Write the list in order

Listing three coins as HHH, HHT, HTT, TTT misses half of them. Branch on the first coin, then the second, and you get all 8.

Drill 7 more on listing the outcomes for up to three coins

Concept 2 of 5: Exactly k heads from n coins: choose the positions

With six coins you cannot list 64 outcomes quickly. Count instead how many ways there are to choose which coins show heads. Exactly 2 heads from 6 coins means choosing 2 of the 6 places:

^6C_2 = 15

ways out of 64.

Definition

$^nC_k = \dfrac{n!}{k!\,(n - k)!}$ is the number of ways to choose k things from n. $^nC_0 = {}^nC_n = 1$ , $^nC_1 = n$ , and $^nC_k = {}^nC_{n - k}$ .
The probability of exactly k heads in n tosses is $\dfrac{^nC_k}{2^n}$ .
Exactly k tails is exactly $n - k$ heads, so it has the same count.
For a range ('at least 3', 'more heads than tails', 'differ by 2'), turn it into a list of exact head counts, then add their counts.
The counts for 4 coins are 1, 4, 6, 4, 1 (total 16); for 6 coins, 1, 6, 15, 20, 15, 6, 1 (total 64).

Exactly k heads in n tosses

P(k \text{ heads}) = \dfrac{^nC_k}{2^n}

Worked example

Six coins are tossed. Find the probability of at least 5 heads.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Probability · Coin Tosses

Five coins are tossed. What is the probability that the number of heads and tails differ by at most

1

Exactly is not at least

'At least 3 heads from 4 coins' means 3 or 4 heads:

\tfrac{4 + 1}{16} = \tfrac{5}{16}

. Stopping at exactly 3 gives

\tfrac{4}{16}

The total is 2 to the power n

Six coins have

2^6 = 64

outcomes, not

2 \times 6 = 12

Drill 15 more on exactly k heads from n coins: choose the positions

Concept 3 of 5: At least one: one minus none

'At least one head' is every outcome except all tails. With 4 coins that is

1 - \tfrac{1}{16} = \tfrac{15}{16}

: one subtraction instead of four cases.

Definition

The probability of at least one head in n tosses is $1 - \left(\tfrac12\right)^n$ .
'At most k' is everything except more than k. At most 3 heads from 4 coins leaves out only HHHH: $1 - \tfrac{1}{16}$ .
If the probability is given, work back: $1 - \left(\tfrac12\right)^n = \tfrac{15}{16}$ gives $\left(\tfrac12\right)^n = \tfrac{1}{16}$ , so $n = 4$ .
A repeated go: two coins tossed together give at least one head with probability $\tfrac34$ . For that to happen on each of k goes, multiply: $\left(\tfrac34\right)^k$ .

At least one

P(\text{at least one}) = 1 - P(\text{none})

Worked example

A coin is tossed 6 times. Find the probability of at least one tail.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Probability · Coin Tosses

A fair coin is tossed

n

times. If the probability of getting at least one head is

\frac{7}{8}

, find

n

None is one outcome

With 4 coins, 'no head' is the single outcome TTTT, probability

\tfrac{1}{16}

. Writing

1 - \tfrac12

for 'at least one' ignores how many coins there are.

On every go means multiply the goes

'At least one head on each of 3 goes of two coins' is

\left(\tfrac34\right)^3 = \tfrac{27}{64}

. The number

1 - \left(\tfrac14\right)^3

answers a different question: a head on at least one of the goes.

Drill 7 more on at least one: one minus none

Concept 4 of 5: Order matters: given tosses, a first head, patterns and a biased coin

Each toss is independent with probability

\tfrac12

, so any one sequence of n tosses has probability

\left(\tfrac12\right)^n

. A question about order is a count of the sequences that fit.

Definition

A condition on particular tosses ignores the rest: 'the first two tosses are heads', out of 6 tosses, is $\left(\tfrac12\right)^2 = \tfrac14$ .
A first head on toss k needs $k - 1$ tails and then a head: probability $\left(\tfrac12\right)^k$ .
Patterns such as alternating faces or heads side by side: list the sequences that fit, then divide by $2^n$ .
'Tossed until' a pattern appears: the pattern must finish exactly on the last toss and must not have appeared earlier.
A biased coin with $P(H) = p$ : multiply each toss's own probability, then add the orders. Exactly one head in two tosses is $2p(1 - p)$ .

One sequence of n tosses

P(\text{a given sequence}) = \left(\tfrac12\right)^n

Worked example

A coin is tossed 5 times. Find the probability that heads and tails alternate.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Probability · Coin Tosses

Four fair coins are tossed. Find the probability that exactly two coins show heads and these two heads are on adjacent coins.

Positions are not counts

'Heads on the first two of 5 tosses' is

\tfrac14

. 'Exactly two heads in 5 tosses' is

\tfrac{10}{32}

, because the two heads can sit in any two places.

The pattern must not finish early

Tossing until two heads come in a row, exactly 3 tosses needs THH, so

\tfrac18

. HHH does not count: it stopped after 2 tosses.

Drill 10 more on order matters: given tosses, a first head, patterns and a biased coin

Concept 5 of 5: Given that: count inside the condition

'Given that' shrinks the sample space to the outcomes that meet the condition. The answer is the favourable outcomes inside it, divided by how many outcomes it has.

Definition

When outcomes are equally likely, $P(A \mid B) = \dfrac{n(A \text{ and } B)}{n(B)}$ . Read it as 'A given B'.
List the outcomes in B first, then count those that are also in A.
The same idea comes back later in the chapter with dice, cards and bags.

Conditional probability

P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}

Worked example

Two coins are tossed. Given that at least one shows a head, find the probability that both show heads.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 5 · Probability · Coin Tosses

Three coins are tossed. What is the probability that the first coin shows head given that exactly two heads appear?

Divide by the condition, not by everything

Given at least one head with two coins, divide by 3, not by 4. Dividing by the whole sample space answers 'both heads', which is

\tfrac14

Drill 2 more on given that: count inside the condition

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Listing the outcomes for up to three coins
Outcomes for n coins
$n(S) = 2^n, \qquad P(E) = \dfrac{n(E)}{2^n}$
Exactly k heads from n coins: choose the positions
Exactly k heads in n tosses
$P(k \text{ heads}) = \dfrac{^nC_k}{2^n}$
At least one: one minus none
At least one
$P(\text{at least one}) = 1 - P(\text{none})$
Order matters: given tosses, a first head, patterns and a biased coin
One sequence of n tosses
$P(\text{a given sequence}) = \left(\tfrac12\right)^n$
Given that: count inside the condition
Conditional probability
$P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$

Watch out for (9)

HT and TH are two outcomes
→ Listing the outcomes for up to three coins
Write the list in order
→ Listing the outcomes for up to three coins
Exactly is not at least
→ Exactly k heads from n coins: choose the positions
The total is 2 to the power n
→ Exactly k heads from n coins: choose the positions
None is one outcome
→ At least one: one minus none
On every go means multiply the goes
→ At least one: one minus none
Positions are not counts
→ Order matters: given tosses, a first head, patterns and a biased coin
The pattern must not finish early
→ Order matters: given tosses, a first head, patterns and a biased coin
Divide by the condition, not by everything
→ Given that: count inside the condition

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