Probability
Basic probability, conditional events
Probability is a branch of mathematics that deals with the likelihood of events occurring. It quantifies uncertainty by assigning a numerical value between 0 and 1 to each possible outcome, where 0 means impossible and 1 means certain. Probability is essential in competitive exams for solving problems involving cards, dice, coins, and real-world scenarios.
Key Formulas
Key Concepts
Classical (Theoretical) Probability
This is the basic definition where probability is calculated as the ratio of favourable outcomes to total possible outcomes, assuming all outcomes are equally likely. For example, P(getting a head) = 1/2 when tossing a fair coin. This approach works well for simple, well-defined experiments like coin tosses, dice rolls, and drawing cards from a well-shuffled deck.
Complementary Events
The complement of an event E (denoted as E' or not E) consists of all outcomes where E does not occur. Since either E happens or it doesn't, P(E) + P(E') = 1. This is useful when calculating P(E) directly is difficult but calculating P(E') is easier. For example, P(at least one head in 3 tosses) = 1 - P(no heads) = 1 - 1/8 = 7/8.
Mutually Exclusive Events
Two events are mutually exclusive (or disjoint) if they cannot occur simultaneously. For example, when rolling a die, getting a 2 and getting a 5 are mutually exclusive. For mutually exclusive events A and B: P(A union B) = P(A) + P(B). This is the addition rule for mutually exclusive events. Note that complementary events are always mutually exclusive, but not all mutually exclusive events are complementary.
Independent Events
Two events are independent if the occurrence of one does not affect the probability of the other. For example, successive coin tosses are independent. For independent events A and B: P(A intersect B) = P(A) x P(B). This is the multiplication rule for independent events. Be careful not to confuse independence with mutual exclusivity - mutually exclusive events with non-zero probability cannot be independent.
Conditional Probability
Conditional probability P(A|B) represents the probability of event A occurring given that event B has already occurred. The formula is P(A|B) = P(A intersect B) / P(B), where P(B) > 0. This concept is crucial for problems involving drawing without replacement, where the probability changes after each draw. For example, P(second card is ace | first card was ace) = 3/51 when drawing from a standard deck.
Cards, Dice, and Coins
Standard deck: 52 cards with 4 suits (spades , hearts , diamonds , clubs ), each with 13 cards (A, 2-10, J, Q, K). Face cards: Jack, Queen, King (12 total). Honour cards: Ace, 10, J, Q, K (20 total). Standard die: 6 faces numbered 1-6. For coins, remember that each toss has 2 outcomes. When multiple coins are tossed, total outcomes = 2^n. For multiple dice, total outcomes = 6^n.
Solved Examples
Problem 1:
A card is drawn from a well-shuffled pack of 52 cards. What is the probability of drawing a king or a queen?
Solution:
Step 1: Identify total outcomes.
Total cards = 52.
Step 2: Identify favourable outcomes.
Number of kings = 4 (one in each suit).
Number of queens = 4 (one in each suit).
Since king and queen are mutually exclusive (a card cannot be both), we add them.
Favourable outcomes = 4 + 4 = 8.
Step 3: Calculate probability.
P(king or queen) = 8/52 = 2/13.
Answer: 2/13
Problem 2:
Two dice are thrown simultaneously. What is the probability of getting a sum of 7?
Solution:
Step 1: Calculate total possible outcomes.
Each die has 6 faces.
Total outcomes = 6 x 6 = 36.
Step 2: Find favourable outcomes (pairs that sum to 7).
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Favourable outcomes = 6.
Step 3: Calculate probability.
P(sum = 7) = 6/36 = 1/6.
Answer: 1/6
Problem 3:
A bag contains 4 red, 5 blue, and 3 green balls. Two balls are drawn at random without replacement. What is the probability that both are red?
Solution:
Step 1: Calculate total balls and probability of first red.
Total balls = 4 + 5 + 3 = 12.
P(first is red) = 4/12 = 1/3.
Step 2: Calculate conditional probability of second red.
After drawing one red ball, 3 red balls remain out of 11 total balls.
P(second is red | first was red) = 3/11.
Step 3: Apply multiplication rule.
P(both red) = P(first red) x P(second red | first red)
P(both red) = (4/12) x (3/11) = 12/132 = 1/11.
Answer: 1/11
Problem 4:
Three coins are tossed simultaneously. What is the probability of getting at least 2 heads?
Solution:
Step 1: Calculate total outcomes.
Total outcomes = 2^3 = 8 (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT).
Step 2: Identify favourable outcomes for 'at least 2 heads'.
This means 2 heads or 3 heads.
2 heads: HHT, HTH, THH (3 outcomes).
3 heads: HHH (1 outcome).
Total favourable = 3 + 1 = 4.
Step 3: Calculate probability.
P(at least 2 heads) = 4/8 = 1/2.
Alternative method using complementary events:
P(at least 2 heads) = 1 - P(0 heads) - P(1 head)
P(0 heads) = 1/8, P(1 head) = 3/8
P(at least 2 heads) = 1 - 1/8 - 3/8 = 4/8 = 1/2.
Answer: 1/2
Tips & Tricks
- Always verify if events are mutually exclusive before using simple addition. Events like 'king' and 'spade' are NOT mutually exclusive (king of spades exists).
- For 'at least one' type problems, use complementary probability: P(at least one) = 1 - P(none). This often simplifies calculations significantly.
- When dealing with cards, remember: P(face card) = 12/52 = 3/13, P(honour card) = 20/52 = 5/13, P(card of a specific suit) = 13/52 = 1/4.
- For problems with replacement, probabilities remain constant. For without replacement, probabilities change and conditional probability applies.
- With dice problems, create a systematic list or table to ensure you count all favourable outcomes correctly, especially for sum-based questions.
- When tossing n coins simultaneously or one coin n times, the probability calculation is identical: each has 2^n equally likely outcomes.
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