The coin toss probability formula helps us find the chance of getting a head or a tail. When you flip a fair coin, there are two possibilities . Each possibility is equally likely. This makes the topic easy to understand for the students. It is often used in mathematics and basic probability problems. Children feel confident comparing possible outcomes and solving basic questions when they know this formula. A coin toss is a good example. Easy, familiar, and fair. It also gives a good basis for later classes in learning more about probability.
A coin toss is used in probability because it is a fair and random experiment. A fair coin has exactly two sides heads (H) and tails (T) and neither side is more likely to appear than the other.

Each outcome has an equal chance of occurring.
Each coin toss is independent. The result of one toss does NOT affect the next toss. This makes it a random experiment you cannot predict the exact result in advance.

The sample space is the set of all possible outcomes.
One coin toss:
Sample Space (S) = {H, T}
Total outcomes = 2
H = Heads
T = Tails
Two coins tossed together:
Sample Space = {HH, HT, TH, TT}
Total outcomes = 4
HH = Both heads
HT = First head, second tail
TH = First tail, second head
TT = Both tails
Outcomes of Three Coin Tosses
Three coins tossed together:
Sample Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Total outcomes = 8
Total outcomes formula: 2ⁿ
Where n = number of coins
1 coin → 2¹ = 2 outcomes
2 coins → 2² = 4 outcomes
3 coins → 2³ = 8 outcomes
Number of favorable outcomes
P(Event) = _____________________________
Total number of possible outcomes
P(E) = n(E) / n(S)
One coin toss:
Favorable outcome = {H} → 1 outcome
Total outcomes = 2
P(Head) = 1/2 = 0.5 = 50%
One coin toss:
Favorable outcome = {T} → 1 outcome
Total outcomes = 2
P(Tail) = 1/2 = 0.5 = 50%
Note: P(H) + P(T) = 1/2 + 1/2 = 1
(All probabilities must sum to 1)
TWO COIN TOSS TABLE:
| Coin 1 | Coin 2 | Outcome |
|---|---|---|
| H | H | HH |
| H | T | HT |
| T | H | TH |
| T | T | TT |
Total outcomes = 4
P(both heads) = P(HH) = 1/4
P(one head, one tail) = P(HT or TH) = 2/4 = 1/2
P(both tails) = P(TT) = 1/4
P(at least one head) = P(HH,HT,TH) = 3/4
P(exactly one head) = P(HT,TH) = 2/4 = 1/2
Sample Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Total outcomes = 8
Outcomes with exactly one head: {HTT, THT, TTH}
Favorable outcomes = 3
P(exactly one head) = 3/8
Outcomes with exactly two heads: {HHT, HTH, THH}
Favorable outcomes = 3
P(exactly two heads) = 3/8
Outcomes with three heads: {HHH}
Favorable outcomes = 1
P(three heads) = 1/8
Three Coin Probability Summary:


Total = 4 branches, each with P = 1/4
Sum = 4 × 1/4 = 1
P(Event) = n(E) / n(S)
Where:
n(E) = number of favorable outcomes
n(S) = total number of outcomes
Total outcomes for n coins = 2ⁿ
Examples:
1 coin: 2¹ = 2
2 coins: 2² = 4
3 coins: 2³ = 8
4 coins: 2⁴ = 16
10 coins: 2¹⁰ = 1024
Question: A fair coin is tossed once. Find the probability of getting a tail.
Solution:
Sample Space = {H, T}
n(S) = 2
Favorable outcome = {T}
n(E) = 1
P(Tail) = 1/2
Answer: P(Tail) = 1/2
Question: Two coins are tossed simultaneously. Find the probability of getting at least one head.
Solution:
Sample Space = {HH, HT, TH, TT}
n(S) = 4
Favorable outcomes = {HH, HT, TH} → 3 outcomes
P(at least one head) = 3/4
Answer: P(at least one head) = 3/4
Question: Three coins are tossed at once. Find the probability of getting exactly two tails.
Solution:
Sample Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
Outcomes with exactly two tails:
{HTT, THT, TTH} → 3 outcomes
P(exactly two tails) = 3/8
Answer: P(exactly two tails) = 3/8
Q1: What is the sample space when one coin is tossed?
Answer: {H, T}
Q2: What is the total number of outcomes when two coins are tossed?
Answer: 4 (= 2²)
Q3: Find P(getting all heads) when 3 coins are tossed.
Answer: 1/8
Q4: What is P(H) + P(T) for a fair coin?
Answer: 1/2 + 1/2 = 1
Q5: A coin is tossed twice. What is P(both tails)?
a) 1/2 b) 1/4 c) 1/3 d) 3/4
Answer: b) 1/4
Q6: Three coins are tossed. P(at least one tail) = ?
a) 1/8 b) 3/8 c) 7/8 d) 5/8
Answer: c) 7/8
(Complement of all heads: 1 − 1/8 = 7/8)
Q7: Total outcomes when 4 coins are tossed?
a) 8 b) 12 c) 16 d) 4
Answer: c) 16 (= 2⁴)
Q8: Ravi and Priya toss a coin to decide who bats first. What is the probability that Ravi wins the toss?
Solution:
P(Ravi wins) = P(getting the chosen side) = 1/2
Answer: 1/2
Q9: A teacher tosses two coins in class. Find P(getting exactly one tail).
Sample Space = {HH, HT, TH, TT}
Exactly one tail = {HT, TH} → 2 outcomes
P = 2/4 = 1/2
Answer: 1/2
The probability of an event is calculated using: Probability = Number of favourable outcomes ÷ Total number of possible outcomes
A coin has two possible outcomes: Head (H) and Tail (T). The probability of getting a head is 1/2.
The probability of getting a tail is 1/2 because there are two equally likely outcomes.
The possible outcomes are:
There are 4 possible outcomes.
There is 1 favourable outcome (HH) out of 4 possible outcomes.
Probability = 1/4
The favourable outcomes are HT and TH.
Probability = 2/4 = 1/2
The total number of possible outcomes when tossing multiple coins is given by:
Total outcomes = 2ⁿ
where n is the number of coins tossed.
Examples:
This formula works because each coin has 2 possible outcomes Head (H) or Tail (T).
To calculate the probability of consecutive coin flips, multiply the probability of each individual flip because each toss is independent.
For a fair coin:
Examples:
In general, the probability of getting a specific sequence of n coin flips is:
Probability = (1/2)ⁿ (for a fair coin).
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