Probability with Coins
Coin toss problems are the most fundamental probability experiments. A fair coin has two equally likely outcomes — Head (H) and Tail (T) — making it the simplest model for probability calculations.
When multiple coins are tossed, the number of outcomes grows as powers of 2. One coin gives 2 outcomes, two coins give 4, three coins give 8, and so on.
Coin problems appear frequently in CBSE Class 10 board exams. They test the ability to list sample spaces, identify favourable outcomes, and apply the probability formula correctly.
What is Probability with Coins?
Definition: A coin toss experiment involves flipping one or more fair coins and recording whether each lands on Head (H) or Tail (T).
Key facts:
- A fair coin has P(H) = P(T) = 1/2.
- Each toss is independent — the result of one toss does not affect another.
- For n coins tossed simultaneously, total outcomes = 2ⁿ.
Sample spaces:
| Experiment | Sample Space | n(S) |
|---|---|---|
| 1 coin | {H, T} | 2 |
| 2 coins | {HH, HT, TH, TT} | 4 |
| 3 coins | {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} | 8 |
Important: When 2 coins are tossed, HT and TH are different outcomes (first coin Head + second coin Tail is different from first coin Tail + second coin Head). The sample space has 4 outcomes, not 3.
Probability with Coins Formula
General Formula for Coin Probability:
P(E) = Number of favourable outcomes / 2ⁿ
Where:
- n = number of coins tossed
- 2ⁿ = total number of outcomes
Number of ways to get exactly k heads in n tosses:
ⁿCₖ = n! / (k! × (n−k)!)
Common results:
- P(exactly 1 head in 2 tosses) = ²C₁/4 = 2/4 = 1/2
- P(at least 1 head in 2 tosses) = 3/4
- P(all heads in 3 tosses) = 1/8
- P(at least 1 head in 3 tosses) = 7/8
Derivation and Proof
Why are there 2ⁿ outcomes for n coins?
- Coin 1 has 2 outcomes: H or T.
- For each outcome of Coin 1, Coin 2 has 2 outcomes.
- By the multiplication principle: 2 × 2 = 4 outcomes for 2 coins.
- For 3 coins: 2 × 2 × 2 = 8 outcomes.
- For n coins: 2 × 2 × ... × 2 (n times) = 2ⁿ.
Counting heads systematically (3 coins):
| Number of Heads | Outcomes | Count |
|---|---|---|
| 0 heads | TTT | 1 = ³C₀ |
| 1 head | HTT, THT, TTH | 3 = ³C₁ |
| 2 heads | HHT, HTH, THH | 3 = ³C₂ |
| 3 heads | HHH | 1 = ³C₃ |
Total: 1 + 3 + 3 + 1 = 8 = 2³. Verified.
Types and Properties
Common types of coin probability questions:
- Exactly k heads: Count outcomes with exactly k H's. Use ⁿCₖ for large n.
- At least k heads: Means k or more heads. Often easier to use: P(at least k) = 1 − P(fewer than k).
- At most k heads: Means k or fewer heads.
- No head (all tails): Only 1 outcome: TTT...T. P = 1/2ⁿ.
- All heads: Only 1 outcome: HHH...H. P = 1/2ⁿ.
Key phrases and their meanings:
| Phrase | Mathematical Meaning | Example (2 coins) |
|---|---|---|
| Exactly one head | Number of H = 1 | {HT, TH} → P = 2/4 = 1/2 |
| At least one head | Number of H ≥ 1 | {HH, HT, TH} → P = 3/4 |
| At most one head | Number of H ≤ 1 | {HT, TH, TT} → P = 3/4 |
| No head | Number of H = 0 | {TT} → P = 1/4 |
Methods
Steps to solve coin probability problems:
- Count the coins: Determine n (number of coins).
- Find total outcomes: n(S) = 2ⁿ.
- Write the sample space (for small n ≤ 3). For larger n, use combinations.
- Identify the event: Translate the question into a precise condition (e.g., "at least 2 heads" means H count ≥ 2).
- Count favourable outcomes: List them or use ⁿCₖ.
- Apply the formula: P(E) = n(E)/n(S).
Shortcut for "at least one" problems:
- P(at least 1 head) = 1 − P(no head) = 1 − P(all tails) = 1 − 1/2ⁿ.
- This is much faster than listing all favourable outcomes.
Common mistakes:
- Listing only 3 outcomes for 2 coins: {0 heads, 1 head, 2 heads}. These are NOT equally likely. The correct sample space has 4 equally likely outcomes.
- Confusing "at least one" with "exactly one".
- Forgetting that tossing 2 coins simultaneously is the same as tossing 1 coin twice — both give 4 outcomes.
Solved Examples
Example 1: One Coin Tossed Once
Problem: A fair coin is tossed once. Find the probability of getting a head.
Solution:
Given:
- S = {H, T}, n(S) = 2
- E = {H}, n(E) = 1
Using P(E) = n(E)/n(S):
- P(H) = 1/2
Answer: P(Head) = 1/2
Example 2: Two Coins — Exactly One Head
Problem: Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Solution:
Given:
- S = {HH, HT, TH, TT}, n(S) = 4
- E = {HT, TH} — exactly one H in each
- n(E) = 2
Using the formula:
- P(exactly 1 head) = 2/4 = 1/2
Answer: P(exactly one head) = 1/2
Example 3: Two Coins — At Least One Head
Problem: Two coins are tossed. Find the probability of getting at least one head.
Solution:
Given:
- S = {HH, HT, TH, TT}, n(S) = 4
Method 1 (direct counting):
- E = {HH, HT, TH} — at least one H
- P = 3/4
Method 2 (complement):
- P(at least 1 H) = 1 − P(no H) = 1 − P(TT) = 1 − 1/4 = 3/4
Answer: P(at least one head) = 3/4
Example 4: Three Coins — All Heads
Problem: Three coins are tossed simultaneously. Find the probability that all three show heads.
Solution:
Given:
- n(S) = 2³ = 8
- E = {HHH} → n(E) = 1
Using the formula:
- P(all heads) = 1/8
Answer: P(HHH) = 1/8
Example 5: Three Coins — Exactly Two Heads
Problem: Three coins are tossed. Find the probability of getting exactly 2 heads.
Solution:
Given:
- S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8
- E (exactly 2 heads) = {HHT, HTH, THH}
- n(E) = 3
Using the formula:
- P(exactly 2 heads) = 3/8
Answer: P(exactly 2 heads) = 3/8
Example 6: Three Coins — At Most One Head
Problem: Three coins are tossed together. Find the probability of getting at most 1 head.
Solution:
Given:
- n(S) = 8
- "At most 1 head" means 0 heads or 1 head.
- 0 heads: {TTT} → 1 outcome
- 1 head: {HTT, THT, TTH} → 3 outcomes
- n(E) = 1 + 3 = 4
Using the formula:
- P(at most 1 head) = 4/8 = 1/2
Answer: P(at most 1 head) = 1/2
Example 7: Three Coins — At Least Two Tails
Problem: Three fair coins are tossed. Find the probability of getting at least 2 tails.
Solution:
Given:
- n(S) = 8
- "At least 2 tails" means 2 tails or 3 tails.
- 2 tails: {HTT, THT, TTH} → 3 outcomes
- 3 tails: {TTT} → 1 outcome
- n(E) = 3 + 1 = 4
Using the formula:
- P(at least 2 tails) = 4/8 = 1/2
Answer: P(at least 2 tails) = 1/2
Example 8: Using Complementary Probability
Problem: Three coins are tossed. Find the probability that not all coins show the same face.
Solution:
Given:
- n(S) = 8
- "All same face" = {HHH, TTT} → n = 2
Using complement:
- P(all same) = 2/8 = 1/4
- P(not all same) = 1 − 1/4 = 3/4
Answer: P(not all same face) = 3/4
Example 9: One Coin Tossed Three Times
Problem: A coin is tossed 3 times. Find the probability that heads appears on the first toss and tails on the remaining two.
Solution:
Given:
- n(S) = 2³ = 8
- E = {HTT} — a specific ordered outcome
- n(E) = 1
Using the formula:
- P(HTT) = 1/8
Answer: P(Head then Tail then Tail) = 1/8
Example 10: Two Coins — Both Show Same Face
Problem: Two coins are tossed. What is the probability that both coins show the same face?
Solution:
Given:
- S = {HH, HT, TH, TT}, n(S) = 4
- "Same face" = {HH, TT} → n(E) = 2
Using the formula:
- P(same face) = 2/4 = 1/2
Answer: P(both same face) = 1/2
Real-World Applications
Decision Making:
- Coin tosses determine which team bats first in cricket or chooses sides in football.
- Fair resolution of disputes — each party has equal probability (1/2) of winning.
Computer Science:
- Random bit generation — Head = 1, Tail = 0. Sequences of coin flips generate random binary numbers.
- Randomised algorithms use coin-flip logic for decision making.
Statistics:
- Coin toss experiments are the simplest example of a Bernoulli trial — the foundation of binomial probability distributions.
Genetics:
- Gender determination in offspring follows a model similar to a fair coin toss (approximately 50-50 probability for male or female).
Key Points to Remember
- A fair coin has two equally likely outcomes: P(H) = P(T) = 1/2.
- For n coins, total outcomes = 2ⁿ.
- 1 coin → 2 outcomes; 2 coins → 4; 3 coins → 8.
- HT and TH are different outcomes — order matters when listing sample space.
- "Exactly k heads" uses direct counting or ⁿCₖ.
- P(at least 1 head) = 1 − (1/2)ⁿ — use complement for efficiency.
- "At least" means ≥, "at most" means ≤, "exactly" means =.
- Tossing 2 coins simultaneously = tossing 1 coin twice. Both produce 4 outcomes.
- Each toss is independent — previous results do not affect future tosses.
- Never list {0 heads, 1 head, 2 heads} as the sample space — these are not equally likely for 2 coins.
Practice Problems
- A coin is tossed once. Find the probability of getting a tail.
- Two coins are tossed simultaneously. Find the probability of getting no head.
- Two coins are tossed. Find P(at most one tail).
- Three coins are tossed together. Find the probability of getting exactly one tail.
- Three coins are tossed. Find the probability of getting at least one head.
- A coin is tossed 4 times. Find the total number of outcomes and the probability of getting all heads.
- Two coins are tossed. Find the probability that the outcomes on both coins are different.
- Three coins are tossed simultaneously. Find the probability of getting more heads than tails.
Frequently Asked Questions
Q1. How many outcomes are there when 2 coins are tossed?
There are 4 outcomes: HH, HT, TH, TT. Each has probability 1/4. Note that HT and TH are separate outcomes because the coins are distinguishable (first coin and second coin).
Q2. What is the probability of getting at least one head when 3 coins are tossed?
P(at least 1 head) = 1 − P(no head) = 1 − P(TTT) = 1 − 1/8 = 7/8.
Q3. Is tossing 2 coins the same as tossing 1 coin twice?
Yes. Both give the same sample space: {HH, HT, TH, TT} with 4 equally likely outcomes. The probability calculations are identical.
Q4. Why are HT and TH considered different outcomes?
Because they represent different results for each coin. In HT, the first coin shows Head and the second shows Tail. In TH, it is reversed. If you treated them as the same, the sample space {HH, HT/TH, TT} would have 3 outcomes that are NOT equally likely (the middle one is twice as probable), leading to wrong calculations.
Q5. What does 'at most 2 heads' mean when tossing 3 coins?
It means 0, 1, or 2 heads. The only excluded case is 3 heads (HHH). So P(at most 2 heads) = 1 − P(3 heads) = 1 − 1/8 = 7/8.
Q6. What is the probability of getting exactly 2 tails in 3 coin tosses?
Outcomes with exactly 2 tails: {HTT, THT, TTH} → 3 outcomes out of 8 total. P = 3/8.
Q7. Does a previous toss affect the next one?
No. Coin tosses are independent events. Getting Head on the first toss does not change the probability of Head on the second toss — it remains 1/2. This is called the independence property.
Q8. How do I solve coin problems for CBSE board exams?
Step 1: Count the coins (n). Step 2: Total outcomes = 2ⁿ. Step 3: List the sample space (for n ≤ 3). Step 4: Count favourable outcomes. Step 5: Apply P(E) = n(E)/2ⁿ. For 'at least' problems, use the complement shortcut.
Q9. Can coin probability be used for biased coins?
The formula P(E) = n(E)/n(S) applies only to fair (unbiased) coins where P(H) = P(T) = 1/2. For biased coins, you need to use the individual probabilities directly and multiply for multiple tosses.
Q10. What is the probability of getting no tail when 3 coins are tossed?
No tail means all heads: E = {HHH}. P = 1/8. This is also equal to (1/2)³ = 1/8.
