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Probability

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1. Basic Definition

Probability is the measure of the likelihood of an event occurring.

P(E)=Number┬аof┬аFavourable┬аOutcomesTotal┬аNumber┬аof┬аPossible┬аOutcomesP(E) = \frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}} or P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

Where n(S)n(S) is the Sample Space.

Example:

Q: Find the probability of getting a Head when tossing a coin.
Solution: Total Outcomes (SS) = {H, T} тЖТn(S)=2\to n(S)=2.
Favourable (EE) = {H} тЖТn(E)=1\to n(E)=1.
P(H)=1/2P(H) = 1/2.

2. Limits of Probability

The probability of an event lies between 0 and 1 (inclusive).

0тЙдP(E)тЙд10 \le P(E) \le 1

  • P(E)=0P(E) = 0: Impossible Event (e.g., Rolling a 7 on a standard die).
  • P(E)=1P(E) = 1: Certain Event (e.g., Sun rising in the East).
  • P(E)+P(not┬аE)=1P(E) + P(\text{not } E) = 1 (Complementary Events).

3. Coins Concepts

  • **1 Coin:** {H, T} $\to 2^1 = 2$.
  • **2 Coins:** {HH, HT, TH, TT} $\to 2^2 = 4$.
  • **3 Coins:** {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} $\to 2^3 = 8$.

**Trick:** For 'At least 1 Head', use $1 - P(\text{No Head})$.

Example:

Q: 3 coins are tossed. Prob of getting at least 1 Head?
Solution: Total = 8. No Head (TTT) = 1 case.
P(At┬аleast┬а1┬аH)=1тИТ1/8=7/8P(\text{At least 1 H}) = 1 - 1/8 = 7/8.

4. Dice Concepts

  • **1 Die:** {1, 2, 3, 4, 5, 6} $\to 6^1 = 6$.
  • **2 Dice:** (1,1) to (6,6) $\to 6^2 = 36$.

**Sum on 2 Dice (Shortcut):**
Min Sum 2 (1 way), Max Sum 12 (1 way).
Sum 7 has max ways (6 ways).
Pattern: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 (for sums 2 to 12).

5. Playing Cards

Total Cards = 52.

  • Suits (13 each): Spades (тЩа), Clubs (тЩг) [Black]; Hearts (тЩе), Diamonds (тЩж) [Red].
  • Face Cards: King, Queen, Jack (3 per suit ├Ч\times 4 = 12 total).
  • Honours: A, K, Q, J.

6. Balls & Marbles (AND/OR Rules)

  • **AND Rule (Multiplication):** Event A AND Event B occur $\to P(A) \times P(B)$. (Independent events).
  • **OR Rule (Addition):** Event A OR Event B occur $\to P(A) + P(B)$ (Mutually Exclusive).

**Selection:** Use Combinations (${}^{n}C_{r}$).
$P(\text{Selection}) = \frac{\text{Favourable Combinations}}{\text{Total Combinations}}$.