Probability, Permutation and Combination

Probability

Probability is a branch of mathematics that deals with the occurrence of random events. It is a measure of the likelihood that an event will occur. The value of probability is always between 0 and 1, inclusive. A probability of 0 means an event is impossible, while a probability of 1 means an event is certain to occur.

Basic Concepts

Experiment: An action or process that produces an outcome. For example, tossing a coin, rolling a die, or drawing a card from a deck.

Sample Space (S): The set of all possible outcomes of an experiment.

• When a coin is tossed, S = {H, T} (H for Heads, T for Tails). The total number of outcomes is |S| = 2.
• When a die is rolled, S = {1, 2, 3, 4, 5, 6}. The total number of outcomes is |S| = 6.
• When two coins are tossed, S = {HH, HT, TH, TT}. The total number of outcomes is |S| = 4.
• When two dice are rolled, S = {(1,1), (1,2), ..., (6,6)}. The total number of outcomes is |S| = 6 * 6 = 36.

Event (E): A subset of the sample space, representing a specific outcome or a set of outcomes we are interested in.

• If we toss a coin, the event of getting heads is E = {H}.
• If we roll a die, the event of getting an even number is E = {2, 4, 6}.

Number of favorable outcomes: The number of outcomes in the event E, denoted by |E|.

Formula for Probability

The probability of an event E occurring is given by the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely.

P(E) = (Number of favorable outcomes) / (Total number of possible outcomes) = |E| / |S|

Types of Events

• Equally Likely Events: Events that have the same chance of occurring. For example, in a fair coin toss, getting heads is equally likely as getting tails.
• Mutually Exclusive Events: Events that cannot occur at the same time. For example, when rolling a die, getting a 1 and getting a 6 are mutually exclusive events.
• Independent Events: The occurrence of one event does not affect the probability of another event occurring. For example, tossing a coin twice. The outcome of the first toss does not influence the outcome of the second toss.
• Dependent Events: The occurrence of one event affects the probability of another event occurring. For example, drawing two cards from a deck without replacement. The probability of drawing the second card depends on what the first card was.

Addition Rule of Probability

For any two events A and B, the probability that either A or B occurs (or both) is given by:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Where:

• P(A ∪ B) is the probability of A or B or both occurring.
• P(A) is the probability of A occurring.
• P(B) is the probability of B occurring.
• P(A ∩ B) is the probability of both A and B occurring.

If events A and B are mutually exclusive (they cannot happen together), then P(A ∩ B) = 0, and the formula simplifies to:

P(A ∪ B) = P(A) + P(B)

Multiplication Rule of Probability

For two events A and B, the probability that both A and B occur is given by:

P(A ∩ B) = P(A) * P(B|A)

Where P(B|A) is the conditional probability of event B occurring given that event A has already occurred.

If events A and B are independent, then P(B|A) = P(B), and the formula simplifies to:

P(A ∩ B) = P(A) * P(B)

Conditional Probability

The probability of event B occurring given that event A has already occurred is denoted as P(B|A).

P(B|A) = P(A ∩ B) / P(A), provided P(A) > 0.

Examples

Example 1: A bag contains 5 red balls and 3 blue balls. If a ball is drawn at random, what is the probability that it is red?

• Total number of balls = 5 (red) + 3 (blue) = 8
• Number of red balls (favorable outcomes) = 5
• Probability of drawing a red ball = 5 / 8

Example 2: Two dice are rolled. What is the probability of getting a sum of 7?

• Total possible outcomes = 6 * 6 = 36
• Favorable outcomes (pairs that sum to 7): (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 such pairs.
• Probability of getting a sum of 7 = 6 / 36 = 1 / 6

Example 3: A card is drawn from a standard deck of 52 cards. What is the probability of drawing a King or a Queen?

• Total cards = 52
• Number of Kings = 4
• Number of Queens = 4
• The events 'drawing a King' and 'drawing a Queen' are mutually exclusive.
• P(King) = 4/52
• P(Queen) = 4/52
• P(King or Queen) = P(King) + P(Queen) = 4/52 + 4/52 = 8/52 = 2/13

Example 4: A card is drawn from a standard deck of 52 cards and not replaced. Then a second card is drawn. What is the probability that both cards are Aces?

• Total cards = 52
• Number of Aces = 4
• Probability of drawing an Ace on the first draw = 4/52
• After drawing one Ace, there are 51 cards left, and 3 of them are Aces.
• Probability of drawing a second Ace given the first was an Ace = 3/51
• Probability of drawing two Aces = (4/52) * (3/51) = 12 / 2652 = 1 / 221

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Permutation and Combination

Permutation and Combination are methods used to count the number of ways in which a set of objects can be arranged or selected. They are fundamental in probability and combinatorics. The key difference lies in whether the order of selection matters.

Permutation

A permutation is an arrangement of objects in a specific order. The order in which the objects are arranged is important.

Formula for Permutation: The number of permutations of 'n' distinct objects taken 'r' at a time is denoted by P(n, r) or ⁿP_r.

ⁿP_r = n! / (n-r)!

Where '!' denotes the factorial. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. By definition, 0! = 1.

Example: How many ways can the letters A, B, C be arranged?

• Here, n=3 (total letters) and r=3 (letters to be arranged).
• ³P₃ = 3! / (3-3)! = 3! / 0! = 6 / 1 = 6.
• The arrangements are: ABC, ACB, BAC, BCA, CAB, CBA.

Example: How many ways can the first, second, and third prizes be awarded to 10 contestants?

• Here, n=10 (contestants) and r=3 (prizes). The order matters (first prize is different from second).
• ¹⁰P₃ = 10! / (10-3)! = 10! / 7! = 10 × 9 × 8 = 720.

Permutation with Repetition: If there are 'n' objects where there are n₁ identical objects of type 1, n₂ identical objects of type 2, ..., n_k identical objects of type k, then the number of distinct permutations is:

n! / (n₁! * n₂! * ... * n_k!)

Example: How many distinct permutations of the letters in the word 'MISSISSIPPI'?

• Total letters (n) = 11
• M = 1 (n₁=1)
• I = 4 (n₂=4)
• S = 4 (n₃=4)
• P = 2 (n₄=2)
• Number of permutations = 11! / (1! * 4! * 4! * 2!) = 39,916,800 / (1 * 24 * 24 * 2) = 39,916,800 / 1152 = 34,650.

Combination

A combination is a selection of objects where the order does not matter. We are only concerned with which objects are chosen, not the order in which they are chosen.

Formula for Combination: The number of combinations of 'n' distinct objects taken 'r' at a time is denoted by C(n, r), ⁿC_r, or $\binom{n}{r}$.

ⁿC_r = n! / [r! * (n-r)!]

Notice that ⁿC_r = ⁿP_r / r!. This is because for every combination of 'r' items, there are r! ways to arrange them (which is what permutation counts).

Example: From a group of 5 people, how many ways can a committee of 3 be selected?

• Here, n=5 (total people) and r=3 (committee size). The order of selection doesn't matter for a committee.
• ⁵C₃ = 5! / [3! * (5-3)!] = 5! / (3! * 2!) = (5 × 4 × 3 × 2 × 1) / [(3 × 2 × 1) × (2 × 1)] = 120 / (6 × 2) = 120 / 12 = 10.
• There are 10 ways to select a committee of 3 from 5 people.

Example: A team of 4 players needs to be selected from 6 players. How many different teams can be formed?

• Here, n=6 and r=4.
• ⁶C₄ = 6! / [4! * (6-4)!] = 6! / (4! * 2!) = (6 × 5 × 4 × 3 × 2 × 1) / [(4 × 3 × 2 × 1) × (2 × 1)] = 720 / (24 × 2) = 720 / 48 = 15.
• There are 15 different teams that can be formed.

Key Property of Combinations: ⁿC_r = ⁿC_(n-r). This is useful for simplifying calculations. For example, ¹⁰C₈ = ¹⁰C_(10-8) = ¹⁰C₂.

Relationship between Permutation and Combination

ⁿP_r = ⁿC_r * r!

This means the number of permutations is equal to the number of combinations multiplied by the number of ways to arrange the selected items.

Applications in Probability

Permutations and Combinations are crucial for calculating probabilities when dealing with selections or arrangements.

Example: A committee of 3 is to be selected from a group of 5 men and 4 women. What is the probability that the committee consists of exactly 2 men and 1 woman?

• Total number of people = 5 men + 4 women = 9.
• Total ways to select a committee of 3 from 9 people (n=9, r=3): ⁹C₃ = 9! / (3! * 6!) = (9 × 8 × 7) / (3 × 2 × 1) = 3 × 4 × 7 = 84. This is our total possible outcomes.
• Ways to select exactly 2 men from 5 men (n=5, r=2): ⁵C₂ = 5! / (2! * 3!) = (5 × 4) / (2 × 1) = 10.
• Ways to select exactly 1 woman from 4 women (n=4, r=1): ⁴C₁ = 4! / (1! * 3!) = 4.
• Number of ways to form a committee with 2 men and 1 woman = ⁵C₂ × ⁴C₁ = 10 × 4 = 40. This is our number of favorable outcomes.
• Probability = (Favorable Outcomes) / (Total Outcomes) = 40 / 84 = 10 / 21.

Example: Four cards are drawn from a standard deck of 52 cards. What is the probability that all four cards are of the same suit?

• Total number of ways to draw 4 cards from 52 (n=52, r=4): ⁵²C₄ = 52! / (4! * 48!) = (52 × 51 × 50 × 49) / (4 × 3 × 2 × 1) = 13 × 17 × 25 × 49 = 270,725. This is the total possible outcomes.
• There are 4 suits (Hearts, Diamonds, Clubs, Spades), each with 13 cards.
• Ways to choose 4 cards from Hearts (n=13, r=4): ¹³C₄ = 13! / (4! * 9!) = (13 × 12 × 11 × 10) / (4 × 3 × 2 × 1) = 13 × 11 × 5 = 715.
• Since there are 4 suits, the total number of ways to get 4 cards of the same suit is 4 × ¹³C₄ = 4 × 715 = 2860. This is the number of favorable outcomes.
• Probability = (Favorable Outcomes) / (Total Outcomes) = 2860 / 270725.
• Simplify the fraction: 2860 / 270725 = (4 × 715) / (715 × 378.63) ... let's simplify by common factors. Both are divisible by 5: 572 / 54145. Both are divisible by 13: 44 / 4165. The probability is 44/4165.