Complete Class 9 Maths Chapter 7 notes on experimental probability, events, trials, and solved examples.
Probability is a mathematical measurement used to quantify the likelihood or chance of an event occurring. Just as length measures physical distance and area measures surface space, probability measures uncertainty.
A Random Experiment is an action or observation that can be repeated under identical conditions, where you know all possible outcomes, but cannot predict which outcome will occur in any single trial (e.g., tossing a coin, rolling a die, selecting a slip in a lucky draw).
Why is Weather or Rain Random? Weather depends on thousands of sensitive atmospheric factors (humidity, wind speed, pressure, temperature). Small shifts produce unpredictable individual daily results, but historical weather data allows us to assign objective probabilities!
The probability of any event \( E \), denoted by \( P(E) \), is always measured on a scale strictly between 0 and 1 (or 0% to 100%):
Rank the following events on the 0 to 1 scale:
1. Next Monday comes after Sunday: Certain (\( P = 1 \)).
2. Snowfall in Mumbai in July: Impossible (\( P = 0 \)).
3. Getting Heads on a fair coin toss: Even Chance (\( P = 0.5 \)).
There are two main objective approaches to estimating probability:
Based on actual collected evidence or data from repeated trials:
\[ \text{Experimental } P(E) = \frac{\text{Number of times event occurred}}{\text{Total number of trials}} \]
Based on logical reasoning assuming all outcomes are equally likely:
\[ \text{Theoretical } P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \]
Problem: You roll a 6-sided die 50 times, and it lands on a '4' exactly 8 times.
(A) What is the experimental probability of rolling a 4?
(B) What is the theoretical probability of rolling a 4?
Solution:
(A) \( \text{Experimental } P(4) = \frac{\text{Times landed on 4}}{\text{Total rolls}} = \frac{8}{50} = \mathbf{0.16 \text{ (or 16\%)}} \).
(B) \( \text{Theoretical } P(4) = \frac{\text{Favourable outcome (1)}}{\text{Total possible outcomes (6)}} = \frac{1}{6} \approx \mathbf{0.167 \text{ (or 16.7\%)}} \).
In real life, surveying an entire population (e.g., 1500 students in a school) is impractical. Analysts collect data from a representative sample (e.g., 50 students). If 20 out of 50 sample students prefer mangoes (\( P = 0.4 \)), we project that \( 0.4 \times 1500 = 600 \) mangoes should be ordered for the school.
For a small number of trials (e.g., 10 rolls), experimental probability may differ significantly from theoretical probability. However, as the number of trials increases to 100, 1,000, or 10,000, experimental probability converges closer and closer to theoretical probability!
Many people mistakenly believe that if a coin lands on Heads 6 times in a row, Tails is "due" on the next toss. Truth: The coin has NO memory! Each toss or die roll is an independent event. On the 7th toss, the probability of Tails remains exactly 50% (\( 1/2 \))!
To analyze probability mathematically, we construct two essential sets:
The complete set of ALL possible outcomes of a random experiment. The size of this set is \( n(S) \).
Any specific outcome or combination of outcomes. An event is a subset of the sample space (\( E \subseteq S \)).
| Experiment | Sample Space (\( S \)) | Sample Size \( n(S) \) |
|---|---|---|
| Tossing 1 Fair Coin | \( S = \{H, T\} \) | \( n(S) = 2 \) |
| Rolling 1 Standard Die | \( S = \{1, 2, 3, 4, 5, 6\} \) | \( n(S) = 6 \) |
| Tossing 2 Coins Simultaneously | \( S = \{HH, HT, TH, TT\} \) | \( n(S) = 4 \) |
| Tossing 3 Coins Simultaneously | \( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \) | \( n(S) = 8 \) |
Problem: Two coins are tossed simultaneously. Find the probability of getting:
(A) At least one Head.
(B) Exactly two Heads.
Solution:
Sample Space \( S = \{HH, HT, TH, TT\} \), so total outcomes \( n(S) = 4 \).
(A) Event "At least 1 Head" \( E_1 = \{HH, HT, TH\} \implies n(E_1) = 3 \).
\[ P(E_1) = \frac{3}{4} = \mathbf{0.75 \text{ (or 75\%)}} \]
(B) Event "Exactly 2 Heads" \( E_2 = \{HH\} \implies n(E_2) = 1 \).
\[ P(E_2) = \frac{1}{4} = \mathbf{0.25 \text{ (or 25\%)}} \]
A Tree Diagram is a visual branching map used to display all outcomes of multi-step experiments (such as tossing a coin twice or picking fruit from two different baskets).
Problem: Basket A contains 1 Apple (A) and 2 Oranges (O₁, O₂). Basket B contains 1 Banana (B) and 1 Mango (M). You pick 1 fruit from each basket.
(i) List the sample space using a tree diagram.
(ii) What is the probability of picking 1 Apple and 1 Banana?
Solution:
Tree diagram branches:
• From Basket A: Choice of Apple (A), Orange 1 (O₁), or Orange 2 (O₂).
• From Basket B: Each splits into Banana (B) or Mango (M).
Sample Space \( S = \{(A, B), (A, M), (O_1, B), (O_1, M), (O_2, B), (O_2, M)\} \implies n(S) = 6 \).
Favourable outcome for (Apple, Banana) = \( \{(A, B)\} \implies n(E) = 1 \).
\[ P(\text{Apple and Banana}) = \frac{1}{6} \approx \mathbf{0.167 \text{ (or 16.7\%)}} \]