Chapter 14: Probability

Class 10 Mathematics - Chapter 14: Probability
Chapter 14: Probability
1. Theoretical (Classical) Probability
Theoretical Probability: The probability of an event $E$, written as $P(E)$, is defined as: $$P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes of the experiment}}$$
Key Axioms & Properties:
  • Impossible Event: An event that cannot happen has $P(E) = 0$.
  • Sure / Certain Event: An event that is guaranteed to happen has $P(E) = 1$.
  • Probability Range: For any event $E$: $$0 \le P(E) \le 1$$
  • Complementary Events: If $\bar{E}$ represents "not $E$": $$P(E) + P(\bar{E}) = 1 \implies P(\bar{E}) = 1 - P(E)$$
  • Sum of Elementary Events: The sum of the probabilities of all the elementary events of an experiment is equal to $1$.
0 Impossible 0.5 Equally Likely 1 Certain
2. Common Random Experiments
1. Tossing Coins:
  • 1 Coin: Sample Space $S = \{H, T\} \implies n(S) = 2$
  • 2 Coins: Sample Space $S = \{HH, HT, TH, TT\} \implies n(S) = 4$
  • 3 Coins: Sample Space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \implies n(S) = 8$

2. Rolling Dice:
  • 1 Die: $S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6$
  • 2 Dice: Total possible outcomes $n(S) = 6 \times 6 = 36$. Pairs range from $(1,1)$ to $(6,6)$.
3. Standard Deck of Cards (52 Cards):
Color Suits Cards per Suit Face Cards (per suit)
Red (26) Hearts ($\heartsuit$), Diamonds ($\diamondsuit$) 13 each Jack, Queen, King (6 Red Total)
Black (26) Spades ($\spadesuit$), Clubs ($\clubsuit$) 13 each Jack, Queen, King (6 Black Total)
3. Solved Examples
Example 3.1: Basic Complementary Probability

If $P(E) = 0.05$, what is the probability of 'not $E$'?

Solution:
We know that: $$P(E) + P(\bar{E}) = 1$$ $$0.05 + P(\bar{E}) = 1$$ $$P(\bar{E}) = 1 - 0.05$$ $$P(\bar{E}) = 0.95$$ The probability of 'not $E$' is $0.95$.

Example 3.2: Card Problems

One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will be: (i) an ace, (ii) not an ace.

Solution:
Total number of possible outcomes $n(S) = 52$.

(i) Let $A$ be the event of getting an ace.
Number of aces in a deck = $4$. $$P(A) = \frac{4}{52} = \frac{1}{13}$$
(ii) Let $\bar{A}$ be the event of not getting an ace. $$P(\bar{A}) = 1 - P(A)$$ $$P(\bar{A}) = 1 - \frac{1}{13} = \frac{12}{13}$$ The probability of drawing an ace is $\frac{1}{13}$ and not an ace is $\frac{12}{13}$.

Example 3.3: Two Dice Rolling

Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is $8$?

Solution:
Total number of outcomes when two dice are thrown = $36$.
The outcomes favourable to the event "sum is 8" are: $$E = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}$$ Number of favourable outcomes $n(E) = 5$.

Using the probability formula: $$P(E) = \frac{n(E)}{n(S)} = \frac{5}{36}$$ The probability that the sum is $8$ is $\frac{5}{36}$.