Probability - Introduction

  • Probability is the measure of the likelihood that an event will occur.
  • It is expressed as a number between 0 and 1, where 0 represents an impossible event and 1 represents a certain event.
  • Probability is widely used in various fields such as statistics, economics, physics, and more.

Probability - Terminology

  • Experiment: An action or process that results in an outcome.
  • Sample Space (S): The set of all possible outcomes of an experiment.
  • Event: A subset of the sample space. It represents a particular outcome or a combination of outcomes.
  • Probability of an Event (P(A)): The likelihood of an event A occurring.

Probability - Types

  • Theoretical Probability: Based on reasoning or logical analysis without conducting an actual experiment.
  • Experimental Probability: Determined through conducting experiments and observing the outcomes.
  • Axiomatic Probability: Founded on mathematical principles and axioms.

Probability - Addition Rule

  • The Addition Rule states that the probability of the union of two events A and B is given by:

    P(A or B) = P(A) + P(B) - P(A and B)

  • P(A and B) represents the probability of both events A and B occurring simultaneously.

Probability - Multiplication Rule

  • The Multiplication Rule states that the probability of the intersection of two independent events A and B is given by:

    P(A and B) = P(A) × P(B)

  • If the events are dependent, the formula becomes:

    P(A and B) = P(A) × P(B|A)

Probability - Complementary Rule

  • The Complementary Rule states that the probability of an event A not occurring is given by:

    P(A’) = 1 - P(A)

  • Here, A’ represents the complement of event A.

Probability - Conditional Probability

  • Conditional Probability is the probability of an event occurring given that another event has already occurred.

  • It is denoted as P(A|B), where A and B are events.

  • The formula for conditional probability is given as:

    P(A|B) = P(A and B) / P(B)

Probability - Bayes’ Theorem

  • Bayes’ Theorem allows us to update the probability of an event based on new information.

  • It is given by the formula:

    P(A|B) = (P(B|A) × P(A)) / P(B)

  • Bayes’ Theorem is widely used in data analysis, statistics, and machine learning.

Probability - Example 1

  • A fair six-sided die is rolled. Find the probability of getting an even number.

    • Sample Space (S) = {1, 2, 3, 4, 5, 6}
    • Event A = {2, 4, 6}
    • Number of favorable outcomes = 3
    • Number of possible outcomes = 6
    • P(A) = 3/6 = 1/2

Probability - Example 2

  • A bag contains 4 red balls and 6 blue balls. A ball is drawn at random. Find the probability of drawing a red ball.

    • Sample Space (S) = {R1, R2, R3, R4, B1, B2, B3, B4, B5, B6}
    • Event A = {R1, R2, R3, R4}
    • Number of favorable outcomes = 4
    • Number of possible outcomes = 10
    • P(A) = 4/10 = 2/5

Probability - Problem 1: Coin Toss

  • A fair coin is tossed. Find the probability of getting heads.
  • Solution:
    • Sample Space (S) = {H, T}
    • Event A = {H}
    • Number of favorable outcomes = 1
    • Number of possible outcomes = 2
    • P(A) = 1/2

Probability - Problem 2: Deck of Cards

  • A standard deck of cards contains 52 cards. Find the probability of drawing a spade.
  • Solution:
    • Sample Space (S) = {all 52 cards}
    • Event A = {spade cards}
    • Number of favorable outcomes = 13 (since there are 13 spade cards)
    • Number of possible outcomes = 52
    • P(A) = 13/52 = 1/4

Probability - Problem 3: Dice Roll

  • Two fair six-sided dice are rolled. Find the probability of getting a sum of 7.
  • Solution:
    • Sample Space (S) = {(1, 1), (1, 2), …, (6, 6)}
    • Event A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
    • Number of favorable outcomes = 6
    • Number of possible outcomes = 36
    • P(A) = 6/36 = 1/6

Probability - Problem 4: Ball Problem

  • A bag contains 8 red balls and 5 blue balls. Two balls are drawn at random without replacement. Find the probability of drawing one red ball and one blue ball.
  • Solution:
    • Sample Space (S) = {all possible pairs of two balls}
    • Event A = {one red ball and one blue ball}
    • Number of favorable outcomes = 8C1 * 5C1 (choosing one red ball from 8 and one blue ball from 5)
    • Number of possible outcomes = 13C2 (choosing any two balls from 13)
    • P(A) = (8C1 * 5C1) / (13C2)

Probability - Problem 5: Three Cards

  • A deck of cards contains 52 cards. Three cards are drawn at random without replacement. Find the probability of drawing three aces.
  • Solution:
    • Sample Space (S) = {all possible sets of three cards}
    • Event A = {three aces}
    • Number of favorable outcomes = 4C3 (choosing three aces from four)
    • Number of possible outcomes = 52C3 (choosing any three cards from 52)
    • P(A) = (4C3) / (52C3)

Probability - Problem 6: Marbles

  • A bag contains 5 red marbles, 3 blue marbles, and 4 green marbles. Three marbles are drawn at random without replacement. Find the probability of drawing two red marbles and one blue marble.
  • Solution:
    • Sample Space (S) = {all possible sets of three marbles}
    • Event A = {two red marbles and one blue marble}
    • Number of favorable outcomes = 5C2 * 3C1 (choosing two red marbles from five and one blue marble from three)
    • Number of possible outcomes = 12C3 (choosing any three marbles from 12)
    • P(A) = (5C2 * 3C1) / (12C3)

Probability - Problem 7: Cards and Suits

  • A standard deck of cards contains 52 cards. If a single card is drawn at random, find the probability of it being a king or a heart.
  • Solution:
    • Sample Space (S) = {all 52 cards}
    • Event A = {king cards} = 4 (as there are 4 kings)
    • Event B = {heart cards} = 13 (as there are 13 hearts)
    • Event A and B = {king of hearts} = 1
    • Number of favorable outcomes = |A ∪ B| = |A| + |B| - |A ∩ B| = 4 + 13 - 1 = 16
    • Number of possible outcomes = 52
    • P(A ∪ B) = 16/52

Probability - Problem 8: Independent Events

  • Two fair coins are tossed. Find the probability of getting heads on the first coin and tails on the second coin.
  • Solution:
    • Sample Space (S) = {HH, HT, TH, TT}
    • Event A = {HH}
    • Event B = {TH}
    • P(A) = 1/4
    • P(B) = 1/4
    • P(A and B) = 1/4
    • P(A) * P(B) = (1/4) * (1/4) = 1/16
    • P(A and B) = P(A) * P(B), so the events are independent.

Probability - Problem 9: Dependent Events

  • A bag contains 3 red balls and 4 blue balls. Two balls are drawn at random without replacement. Find the probability of drawing a red ball followed by a blue ball.
  • Solution:
    • Sample Space (S) = {all possible pairs of two balls}
    • Event A = {red ball on the first draw}
    • Event B = {blue ball on the second draw}
    • Number of favorable outcomes = 3C1 * 4C1 (choosing one red ball from three and one blue ball from four)
    • Number of possible outcomes = 7C2 (choosing any two balls from seven)
    • P(A) = 3/7
    • P(B|A) = 4/6 (since we have already drawn a red ball, there are 6 balls left)
    • P(A and B) = (3/7) * (4/6)

Probability - Problem 10: Digits Combinations

  • A three-digit number is selected at random. Find the probability that the number does not contain the digit 7.
  • Solution:
    • Sample Space (S) = {all three-digit numbers}
    • Event A = {numbers without digit 7}
    • Number of favorable outcomes = 9 * 10 * 10 (as the first digit can be any number from 1 to 9 and the second and third digits can be any number from 0 to 9)
    • Number of possible outcomes = 10 * 10 * 10 (as each digit can be any number from 0 to 9)
    • P(A) = (9 * 10 * 10) / (10 * 10 * 10) = 9/10

Probability - Problem 11: Cards and Numbers

  • A standard deck of cards contains 52 cards. If a single card is drawn at random, find the probability of it being a red card or a number card.
  • Solution:
    • Sample Space (S) = {all 52 cards}
    • Event A = {red cards} = 26 (as there are 26 red cards)
    • Event B = {number cards} = 36 (as there are 36 number cards)
    • Event A and B = {red number cards} = 18 (as there are 18 red number cards)
    • Number of favorable outcomes = |A ∪ B| = |A| + |B| - |A ∩ B| = 26 + 36 - 18 = 44
    • Number of possible outcomes = 52
    • P(A ∪ B) = 44/52

Probability - Problem 12: Dice Roll

  • Two fair six-sided dice are rolled. Find the probability of getting a sum greater than 9.
  • Solution:
    • Sample Space (S) = {(1, 1), (1, 2), …, (6, 6)}
    • Event A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
    • Number of favorable outcomes = 6
    • Number of possible outcomes = 36
    • P(A) = 6/36 = 1/6

Probability - Problem 13: Cards and Suits

  • A standard deck of cards contains 52 cards. If two cards are drawn at random, find the probability that both cards are hearts.
  • Solution:
    • Sample Space (S) = {all possible pairs of two cards}
    • Event A = {two hearts}
    • Number of favorable outcomes = 13C2 (choosing two hearts from thirteen)
    • Number of possible outcomes = 52C2 (choosing any two cards from fifty-two)
    • P(A) = (13C2) / (52C2)

Probability - Problem 14: Marbles

  • A bag contains 5 red marbles, 3 blue marbles, and 4 green marbles. Two marbles are drawn at random without replacement. Find the probability of drawing two marbles of different colors.
  • Solution:
    • Sample Space (S) = {all possible pairs of two marbles}
    • Event A = {two marbles of different colors}
    • Number of favorable outcomes = (5C1 * 3C1) + (5C1 * 4C1) + (3C1 * 4C1) (choosing one red and one blue, or one red and one green, or one blue and one green)
    • Number of possible outcomes = 12C2 (choosing any two marbles from twelve)
    • P(A) = [(5C1 * 3C1) + (5C1 * 4C1) + (3C1 * 4C1)] / (12C2)

Probability - Problem 15: Computer Code

  • In a computer code, there are 10 characters - 4 vowels (a, e, i, o) and 6 consonants (b, c, d, f, g, h). If a character is selected at random, find the probability of it being a vowel or a consonant.
  • Solution:
    • Sample Space (S) = {all 10 characters}
    • Event A = {vowels} = 4 (as there are 4 vowels)
    • Event B = {consonants} = 6 (as there are 6 consonants)
    • Number of favorable outcomes = |A ∪ B| = |A| + |B| = 4 + 6 = 10
    • Number of possible outcomes = 10
    • P(A ∪ B) = 10/10 = 1

Probability - Problem 16: Coin and Dice

  • A fair coin is tossed and a fair six-sided die is rolled. Find the probability of getting heads on the coin and an even number on the die.
  • Solution:
    • Sample Space (S) = {all possible outcomes of a coin and a die}
    • Event A = {heads on the coin}
    • Event B = {even number on the die}
    • Number of favorable outcomes = 1 (getting heads) * 3 (getting an even number) = 3
    • Number of possible outcomes = 2 (coin toss) * 6 (die roll) = 12
    • P(A and B) = 3/12 = 1/4

Probability - Problem 17: Bag of Marbles

  • A bag contains 6 red marbles and 4 blue marbles. Three marbles are drawn at random without replacement. Find the probability of drawing two red marbles followed by a blue marble.
  • Solution:
    • Sample Space (S) = {all possible sets of three marbles}
    • Event A = {two red marbles and one blue marble}
    • Number of favorable outcomes = 6C2 * 4C1 (choosing two red marbles from six and one blue marble from four)
    • Number of possible outcomes = 10C3 (choosing any three marbles from ten)
    • P(A) = (6C2 * 4C1) / (10C3)

Probability - Problem 18: Multiple Choice Questions

  • A multiple-choice question has 4 possible answers, but only one option is correct. If a student randomly guesses the answer, find the probability of guessing the correct answer.
  • Solution:
    • Number of favorable outcomes = 1 (since there is only one correct answer)
    • Number of possible outcomes = 4 (since there are 4 options)
    • P(Correct Answer) = 1/4

Probability - Problem 19: Coin Flips

  • Three fair coins are tossed. Find the probability of getting exactly two heads.
  • Solution:
    • Sample Space (S) = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
    • Event A = {exactly two heads} = {HHT, HTH, THH}
    • Number of favorable outcomes = 3
    • Number of possible outcomes = 8
    • P(A) = 3/8

Probability - Problem 20: Card Game

  • A game is played with a standard deck of cards. If two cards are drawn without replacement, find the probability that the first card is a heart and the second card is an ace.
  • Solution:
    • Sample Space (S) = {all possible pairs of two cards}
    • Event A = {first card is a heart}
    • Event B = {second card is an ace}
    • Number of favorable outcomes = 13 (as there are 13 hearts) * 4 (as there are 4 aces)
    • Number of possible outcomes = 52 (using two cards from a deck