Probability — Class 12 Maths Notes

Topics covered in Probability

1. 1.Introduction of Probability

   Introduction to Probability
   

   Probability is a fascinating area of mathematics that deals with calculating the likelihood of different events. When you flip a coin, for example, you naturally expect it to land on heads or tails. Probability helps us understand and quantify the chances of either outcome.

   What is Probability?

   Probability is the measure of how likely an event is to occur out of the number of possible outcomes. It's expressed as a number between 0 and 1 — with 0 indicating an impossible event, and 1 indicating a certainty. The idea can be easily seen in everyday life; for instance, predicting rain, deciding which team will win a game, or even guessing what color of sock you'll pull out of a drawer blindly.

   How Do We Calculate Probability?

   To calculate the probability of an event happening, you use the formula:

   =Number of favorable outcomesTotal number of outcomesP(E)=Total number of outcomesNumber of favorable outcomes​

   Example in Daily Life:

   Imagine you have a bag with 2 red balls and 3 green balls, and you want to calculate the probability of pulling out a red ball. Here, the total number of outcomes is 5 (since there are 5 balls in total), and the number of favorable outcomes (pulling a red ball) is 2. So, the probability P is:

   (red ball)=25=0.4P(red ball)=52​=0.4

   This means there is a 40% chance that you will pull out a red ball.

   Applications of Probability in Real Life and Careers:

   1. Weather Forecasting: Meteorologists use probability to predict weather patterns like rain, storm, or sunshine. It helps in planning agricultural, water management, and disaster preparedness activities.

   2. Finance and Stock Market: Traders and analysts use probability to predict stock movements and minimize risks. Understanding probability helps in making informed investment decisions.

   3. Health and Medicine: Probability is used in predicting disease outbreaks, understanding the effectiveness of a new drug, or determining the genetic likelihood of inheriting a health condition.

   4. Engineering: Probability is crucial in reliability engineering and quality control. Engineers use it to predict the lifespan of materials or the likelihood of system failures.

   Simple Activity to Understand Probability:

   Try this fun activity: Roll a six-sided dice 30 times and record the outcome each time. Calculate how often each number appears. According to probability theory, each number should appear about 5 times (since 306=5630​=5). This simple experiment helps you understand how probability works in real situations.

   Probability is not just an abstract concept; it's a useful tool that helps us make decisions in uncertain conditions, and it's used widely across many industries to improve processes, predict events, and understand patterns.

2. 2.Conditional Probability

   Conditional Probability

   Conditional probability is a really useful concept in probability theory that comes into play when the outcome of one event affects the outcome of another event. Think about when you check the weather forecast to decide whether to carry an umbrella. If it's already cloudy, the probability of rain might be higher. This is where conditional probability helps us make more informed decisions based on existing conditions.

   What is Conditional Probability?

   Conditional probability measures the probability of an event occurring given that another event has already happened. It's written as (∣)P(A∣B), meaning the probability of event A happening, given that event B has already occurred.

   Properties of Conditional Probability

   1. Non-negativity: Just like any probability, conditional probabilities are always non-negative, meaning they fall between 0 and 1. (∣)≥0P(A∣B)≥0.
   2. Certainty: If event B has occurred, then the probability of B given B is always 1. (∣)=1P(B∣B)=1.
   3. Sum Rule: If 1,2,…,A1​,A2​,…,An​ are mutually exclusive events, then the probability of at least one of them occurring given B is the sum of their individual conditional probabilities:(1∪2∪…∪∣)=(1∣)+(2∣)+…+(∣)P(A1​∪A2​∪…∪An​∣B)=P(A1​∣B)+P(A2​∣B)+…+P(An​∣B)

   Examples to Illustrate Properties

   1. Non-negativity Example:

      • Suppose you have a deck of cards and you're interested in the probability of drawing a king given that a face card has been drawn. Since drawing a king is a subset of drawing a face card, this probability is non-negative.
      • Calculation: There are 12 face cards and 4 kings. So, the probability of drawing a king given that you've drawn a face card is 412=13124​=31​.

   2. Certainty Example:

      • If you already know a coin toss resulted in heads, the probability of heads given heads is 100%.
      • Calculation: (Heads∣Heads)=1P(Heads∣Heads)=1.

   3. Sum Rule Example:

      • Using the same deck of cards, calculate the probability of drawing either a king or a queen given that a face card has been drawn.
      • There are 4 kings and 4 queens among 12 face cards. Since these events are mutually exclusive:(King∪Queen∣Face Card)=(King∣Face Card)+(Queen∣Face Card)=412+412=23P(King∪Queen∣Face Card)=P(King∣Face Card)+P(Queen∣Face Card)=124​+124​=32​

   Applications in Real Life

   Conditional probability isn't just a theoretical concept; it has practical applications across various fields:

   • Weather Forecasting: Meteorologists use it to predict weather conditions given current atmospheric data.
   • Medical Field: Doctors estimate the probability of a disease given certain symptoms or test results.
   • Insurance: Insurers calculate premiums based on the conditional probability of events like accidents or natural disasters.

   Simple Activity to Understand Conditional Probability

   Here’s an easy activity to get a hands-on understanding of conditional probability:

   • Take a pack of playing cards and separate it into reds and blacks.
   • Without looking, choose one card from the entire deck and place it aside. Then shuffle the remaining cards.
   • Now, try to guess the color of the card you set aside, knowing the color of the first card you see from the shuffled deck.
   • Repeat this several times and record the outcomes to see if knowing one outcome helps predict another.

   This approach gives you a practical understanding of how conditional probability works and why it's useful for predicting outcomes based on known conditions. It’s a valuable tool for making more accurate predictions and decisions in everyday situations and professional settings.

3. 3.Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E)

   The conditional probability of an event A given that event B has occurred is denoted by P(A|B) and is given by the formula:

   (∣)=(∩)P(A∣B)=P(B)P(A∩B)​

   where:

   • (∩)P(A∩B) is the probability that both events A and B occur,
   • P(B) is the probability that event B occurs.

   So to find P(E|F), which means the probability of event E occurring given that F has already occurred, we use the given probabilities:

   (∣)=(∩)P(E∣F)=P(F)P(E∩F)​

   Plugging in the given values:

   (∣)=0.20.3P(E∣F)=0.30.2​

   Similarly, to find P(F|E):

   (∣)=(∩)P(F∣E)=P(E)P(E∩F)​

   And we calculate:

   (∣)=0.20.6P(F∣E)=0.60.2​

   Let's compute these values.

   The solutions to the problem are:

   • P(E|F), the probability of event E given that F has occurred, is approximately 0.67.
   • P(F|E), the probability of event F given that E has occurred, is approximately 0.33.

   Now, let's understand this in a real-world context. Suppose event E is "it rains" (with a probability of 0.6), and event F is "you carry an umbrella" (with a probability of 0.3). If the probability that both "it rains" and "you carry an umbrella" is 0.2, then:

   • P(E|F) = 0.67 means that if you've decided to carry an umbrella, the probability of it actually raining is 67%.
   • P(F|E) = 0.33 means that if it is raining, the chance that you've brought an umbrella is 33%.

   This is a simple example of how conditional probability is used in everyday decision-making, such as deciding whether to carry an umbrella based on the weather forecast. In the professional world, similar concepts are used in a vast array of fields including meteorology, insurance, finance, and many forms of data analysis.

4. 4.Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

   To calculate P(A|B), we'll use the formula for conditional probability:

   (∣)=(∩)P(A∣B)=P(B)P(A∩B)​

   This formula tells us how likely event A is to happen assuming that event B has already happened.

   Here's what each part means:

   • P(A|B) is the probability of A given B has occurred.
   • P(A ∩ B) is the probability that both A and B occur together.
   • P(B) is the probability that event B occurs.

   Now, using the values given in your question:

   • P(B) = 0.5
   • P(A ∩ B) = 0.32

   Let's put these numbers into our formula:

   (∣)=0.320.5P(A∣B)=0.50.32​

   When we divide 0.32 by 0.5, we get 0.64.

   So, P(A|B) is 0.64, which means that if event B has occurred, there's a 64% chance that event A will also occur.

5. 5.Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5

   Let's compute the probability (∪)P(A∪B) with the given conditions:

   1. From 2=2P(A)=P(B), we find P(A) given that =513P(B)=135​: =2=5132P(A)=2P(B)​=2135​​

   2. We know (∣)=(∩)P(A∣B)=P(B)P(A∩B)​, so we can rearrange it to find (∩)P(A∩B) given (∣)=25P(A∣B)=52​: (∩)=(∣)×=25×513P(A∩B)=P(A∣B)×P(B)=52​×135​

   3. Finally, we use the formula for the union of two events to find (∪)P(A∪B): (∪)=+−(∩)P(A∪B)=P(A)+P(B)−P(A∩B)

   Let's go ahead and calculate each step:

   For step 1: =513÷2=526P(A)=135​÷2=265​

   For step 2: (∩)=25×513=213P(A∩B)=52​×135​=132​

   For step 3, we plug these into the union formula: (∪)=526+513−213P(A∪B)=265​+135​−132​

   Now we can simplify: (∪)=526+1026−226P(A∪B)=265​+2610​−262​ (∪)=1326−226P(A∪B)=2613​−262​ (∪)=1126P(A∪B)=2611​

   Thus, the probability (∪)P(A∪B) is 11262611​, as expected.

6. 6.Two coins are tossed once: (i) E: a tail appears on one coin, F: one coin shows head , (ii) E: no tail appears, F: no head appears

   Two coins are tossed once:

   (i) E: a tail appears on one coin, F: one coin shows head

   There are four possible outcomes when two coins are tossed: HH, HT, TH, TT.

   • For event E (a tail appears on one coin), the favorable outcomes are HT and TH. Therefore, 𝑃(𝐸)=24=12P(E)=42​=21​.
   • For event F (one coin shows head), the favorable outcomes are HH and HT. Therefore, 𝑃(𝐹)=24=12P(F)=42​=21​.

   (ii) E: no tail appears, F: no head appears

   • For event E (no tail appears), the only favorable outcome is HH. Therefore, 𝑃(𝐸)=14P(E)=41​.
   • For event F (no head appears), the only favorable outcome is TT. Therefore, 𝑃(𝐹)=14P(F)=41​.

7. 7.A die is thrown three times: E: 4 appears on the third toss, F: 6 and 5 appear respectively on the first two tosses

   A die is thrown three times:

   E: 4 appears on the third toss, F: 6 and 5 appear respectively on the first two tosses

   • The outcome of the third toss is independent of the first two tosses. For event E (4 on the third toss), the probability is 𝑃(𝐸)=16P(E)=61​ since there is 1 chance out of 6 to get a 4.
   • For event F (6 and 5 on the first two tosses), the probability is 𝑃(𝐹)=16×16=136P(F)=61​×61​=361​.

8. 8.Mother, father and son line up at random for a family picture: E: son on one end, F: father in middle

   Mother, father and son line up at random for a family picture:

   E: son on one end, F: father in middle

   • There are 3! (or 6) possible arrangements for three people. For event E (son on one end), the son can be either in the first position or the last position. With the other two positions interchangeable, there are 2 possibilities for each end. Therefore, 𝑃(𝐸)=46=23P(E)=64​=32​.
   • For event F (father in middle), there are 2 possible arrangements: mother-son-father and son-mother-father. Therefore, 𝑃(𝐹)=26=13P(F)=62​=31​.

9. 9.Find the conditional probability

   A black and a red dice are rolled. 1. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. 2. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. Solution - 1. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

   When the black die shows a 5, we're interested in the outcomes on the red die that would result in a sum greater than 9. This means the red die needs to show either a 5 or a 6 (because 5 + 5 = 10 and 5 + 6 = 11, both greater than 9). There are 6 possible outcomes for the red die (1 through 6).

   • Favorable outcomes for the red die (to get a sum greater than 9): 5, 6
   • Total possible outcomes for the red die: 6

   The conditional probability, P(sum > 9 | black die = 5), is therefore:

   𝑃(sum > 9 | black die = 5)=Number of favorable outcomesTotal possible outcomes=26P(sum > 9 | black die = 5)=Total possible outcomesNumber of favorable outcomes​=62​

   Solution - 2. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

   If the red die shows a number less than 4, it can be either 1, 2, or 3. We want the total sum to be 8.

   • If the red die is 1, the black die must be 7 for the sum to be 8, which is impossible since a die only has 6 faces.
   • If the red die is 2, the black die must be 6.
   • If the red die is 3, the black die must be 5.

   Since there are only 3 possible outcomes for the red die to be less than 4, the only favorable outcomes that give us a sum of 8 are when the red die is 2 or 3.

   • Favorable outcomes (to get a sum of 8): red die = 2 (black die must be 6), red die = 3 (black die must be 5)
   • Total possible outcomes (where red die < 4): 3

   The conditional probability, P(sum = 8 | red die < 4), is therefore:

   𝑃(sum = 8 | red die < 4)=Number of favorable outcomesTotal possible outcomes=23P(sum = 8 | red die < 4)=Total possible outcomesNumber of favorable outcomes​=32​

   The results are:

   (a) 2662​ or 1331​ for the sum being greater than 9 given a 5 on the black die. (b) 2332​ for the sum being 8 given the red die is less than 4.

10. 10.Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

    Let's Solve the conditional probabilities for both scenarios:

    (i) The youngest is a girl

    When we know the youngest child is a girl, there are only two possible gender combinations for the two children: the oldest could be a boy (BG) or a girl (GG), where G represents a girl and B represents a boy.

    • Favorable outcome (both are girls): GG
    • Total possible outcomes (youngest is a girl): BG, GG

    The conditional probability 𝑃(both girls | youngest girl)P(both girls | youngest girl) is therefore the ratio of the favorable outcome to the total possible outcomes: 𝑃(both girls | youngest girl)=12P(both girls | youngest girl)=21​

    (ii) At least one is a girl

    If at least one child is a girl, there are three possible gender combinations for the two children: GG, BG, GB (where at least one G is in each combination).

    • Favorable outcome (both are girls): GG
    • Total possible outcomes (at least one girl): GG, BG, GB

    The conditional probability 𝑃(both girls | at least one girl)P(both girls | at least one girl) is therefore: 𝑃(both girls | at least one girl)=13P(both girls | at least one girl)=31​

    So the answers are:

    (i) The conditional probability that both children are girls given that the youngest is a girl is 1221​. (ii) The conditional probability that both children are girls given that at least one is a girl is 1331​.

11. 11.An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question ba

    To find the probability that a question is easy given that it is a multiple choice question, we'll use the formula for conditional probability:

    𝑃(Easy∣Multiple Choice)=𝑃(Easy and Multiple Choice)𝑃(Multiple Choice)P(Easy∣Multiple Choice)=P(Multiple Choice)P(Easy and Multiple Choice)​

    Here's what we know:

    • The number of easy multiple choice questions is 500.
    • The total number of multiple choice questions is the sum of easy and difficult multiple choice questions, which is 500 (easy) + 400 (difficult) = 900.
    • We're looking for the probability that a question is easy AND multiple choice, which we already know is 500.

    So, the probability is:

    𝑃(Easy∣Multiple Choice)=500900P(Easy∣Multiple Choice)=900500​

    This simplifies to:

    𝑃(Easy∣Multiple Choice)=59P(Easy∣Multiple Choice)=95​

    So the probability that the question will be easy given that it is a multiple choice question is 5995​.

12. 12.Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

    To solve this, first, we list out all the possible outcomes when two dice are rolled where the results are different:

    These are: (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5).

    This gives us a total of 30 different outcomes where the numbers on the two dice are different.

    Now, for the sum of the numbers to be 4, with the condition that the numbers on both dice are different, we have the following outcomes: (1,3) and (3,1).

    So, we have 2 favorable outcomes out of 30 possible ones.

    The probability P(sum is 4 | numbers are different) is:

    𝑃(sum is 4 | numbers are different)=230P(sum is 4 | numbers are different)=302​

    Simplifying the fraction gives us:

    𝑃(sum is 4 | numbers are different)=115P(sum is 4 | numbers are different)=151​

    So, the probability that the sum of the numbers is 4 given that the two numbers are different is 115151​.

13. 13.Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

    To solve this problem, we need to calculate the conditional probability of the event "the coin shows a tail" given that "at least one die shows a 3." Let's define the events:

    • Event 𝐴A: At least one die shows a 3.
    • Event 𝐵B: The coin shows a tail.

    Step 1: Understand the Sequence of Events

    First, we understand the sequence of events:

    • If a multiple of 3 (i.e., 3 or 6) shows up on the first die throw, the die is thrown again.
    • If any other number (1, 2, 4, or 5) shows up on the first die throw, a coin is tossed.

    Step 2: Calculate Probability of 𝐴A (At least one die shows a 3)

    To calculate 𝑃(𝐴)P(A), consider:

    • First throw is 3 or 6: Continue to throw again.
    • Second throw after getting a 3 or 6: Independent of the first throw if 3 or 6 was rolled.

    Probabilities:

    • First throw is 3 or 6: 26=1362​=31​
    • First throw is not 3 or 6 (a coin is tossed): 46=2364​=32​

    Considering a second die throw (only if the first was 3 or 6, thus a 1331​ probability to re-roll):

    • Probability of getting a 3 at least once if a re-roll happens can happen in two ways:
      • First throw is a 3
      • First throw is a 6, and second throw is a 3

    The total probability of getting a 3, given a multiple of 3 was rolled initially (considering independence of throws): 𝑃(First is 3)+𝑃(First is 6 and Second is 3)=13×12+13×12×16=16+136P(First is 3)+P(First is 6 and Second is 3)=31​×21​+31​×21​×61​=61​+361​

    Step 3: Calculate Probability of 𝐵B given 𝐴A

    • First throw not a multiple of 3 (coin is tossed): Probability the first throw isn't 3 or 6 but results in a coin toss 2332​.
    • Coin toss results in tails: 1221​.

    Now, we need to integrate these probabilities to consider 𝐵B given 𝐴A: 𝑃(𝐵∣𝐴)=𝑃(𝐴 and 𝐵)𝑃(𝐴)P(B∣A)=P(A)P(A and B)​

    Calculating 𝑃(𝐴 and 𝐵)P(A and B)

    • 𝑃(𝐴 and 𝐵)P(A and B) is the probability that at least one die shows a 3 (in all throws) and the coin shows a tail: 𝑃(𝐴 and 𝐵)=𝑃(at least one 3 and tails)P(A and B)=P(at least one 3 and tails)
    • Throwing a 3 and a coin toss happens (and results in tails): (Only if the first throw is not a multiple of 3, and the next throw after a non-multiple of 3 shows a 3, or the first throw is a 3, which results in another throw and finally a 3 that stops further throws).
    • Only possible if first throw is not 3 or 6, next is 3, and then a coin toss results in tails.

    Given the complexity of continuous rerolls, and using simple cases, let's assume we are focusing on the scenario:

    1. First throw not 3 or 6 (so, a coin is tossed)
    2. The coin shows a tail

    𝑃(First not 3 or 6, and tails)=46×12=13P(First not 3 or 6, and tails)=64​×21​=31​

    Then, since 𝑃(𝐴)P(A) involves any scenario where a 3 appears at least once, which includes a multitude of cases due to rerolls and is complex to calculate without more defined rules on stopping conditions, let's approximate:

    Approximate 𝑃(𝐴)P(A) and 𝑃(𝐵∣𝐴)P(B∣A)

    Assuming the chance of eventually getting a 3 given a re-roll is about 16+136=73661​+361​=367​, and using the scenario of a 3 appearing after a non-3/6, which triggers a coin toss:

    𝑃(𝐵∣𝐴)≈13736=127×13=47P(B∣A)≈367​31​​=712​×31​=74​
    
    

14. 14.Multiplication Theorem on Probability

    Introduction to the Multiplication Theorem on Probability

    The Multiplication Theorem on Probability is a fundamental concept that helps us find the probability of two or more events happening in sequence. It tells us how likely it is that two things will both occur. Let's think about this with a simple daily life example: suppose you have a bag of blue and red socks. If you want to know the likelihood of picking two blue socks in a row without looking (and without putting the first sock back), the Multiplication Theorem can help you calculate that probability.

    Understanding the Multiplication Theorem

    The theorem can be divided into two parts:

    1. Independent Events: Two events are independent if the occurrence of one does not affect the occurrence of the other. For independent events, the probability that all events occur is the product of their individual probabilities. Mathematically, if events A and B are independent, then 𝑃(𝐴 and 𝐵)=𝑃(𝐴)×𝑃(𝐵)P(A and B)=P(A)×P(B).

    2. Dependent Events: Two events are dependent if the occurrence of one affects the occurrence of the other. For dependent events, you need to adjust the probability of the second event occurring based on the outcome of the first event. Here, 𝑃(𝐴 and 𝐵)=𝑃(𝐴)×𝑃(𝐵∣𝐴)P(A and B)=P(A)×P(B∣A), where 𝑃(𝐵∣𝐴)P(B∣A) is the probability of B given that A has occurred.

    Real-Life Applications

    This theorem is widely used in various fields such as:

    • Risk Assessment: Businesses use it to calculate the probability of various risks occurring together, like software failure and data breaches.
    • Healthcare: To determine the likelihood of multiple health symptoms leading to a particular diagnosis.
    • Sports: Calculating the probability of a team winning multiple matches in a row.

    Problem Solving Examples

    Let's solve three problems to understand how this theorem works in different scenarios.

    Example 1: Independent Events Imagine you have a dice and a coin. You want to know the probability of rolling a six and then flipping a heads.

    • Step 1: Calculate the probability of rolling a six, 𝑃(six)=16P(six)=61​.
    • Step 2: Calculate the probability of flipping a heads, 𝑃(heads)=12P(heads)=21​.
    • Step 3: Multiply these probabilities (since rolling a dice and flipping a coin are independent events), 𝑃(six and heads)=16×12=112P(six and heads)=61​×21​=121​.

    Example 2: Dependent Events Suppose a bag contains 3 red balls and 2 blue balls. You draw one ball, do not replace it, and then draw another ball. What is the probability of drawing two red balls?

    • Step 1: Probability of drawing the first red ball, 𝑃(red1)=35P(red1)=53​.
    • Step 2: After drawing the first red ball, there are now 4 balls left (2 red, 2 blue). The probability of drawing a second red ball, 𝑃(red2|red1)=24=12P(red2|red1)=42​=21​.
    • Step 3: Multiply these probabilities (since the events are dependent), 𝑃(two reds)=35×12=310P(two reds)=53​×21​=103​.

    Example 3: Mixed Events A locker requires a 3-digit code which is selected by rolling a die three times. What is the probability that the first digit is 5, and the other two digits are different from each other?

    • Step 1: Probability that the first roll gives a 5, 𝑃(first is 5)=16P(first is 5)=61​.
    • Step 2: Probability that the second roll is not 5 (hence different from the first), 𝑃(second not 5)=56P(second not 5)=65​.
    • Step 3: Probability that the third roll is different from the second. Since there are 4 digits left (excluding the digit from the second roll), 𝑃(third different from second)=46=23P(third different from second)=64​=32​.
    • Step 4: Multiply all these probabilities, 𝑃(5 and different)=16×56×23=10108=554P(5 and different)=61​×65​×32​=10810​=545​.

    These examples show how the Multiplication Theorem can be used in various contexts to compute probabilities for both dependent and independent events.

15. 15.Introduction to Independent Events

    Introduction to Independent Events
    

    Imagine you're flipping a coin and rolling a die at the same time. Whether the coin shows heads or tails doesn't affect the number on the die, right? This is a simple example of independent events in probability. Independent events are those where the outcome of one event does not influence the outcome of another.

    Definitions of Independent Events

    1. Statistical Definition: Two events, A and B, are independent if the probability of both events occurring together (A and B) is the product of their individual probabilities:

       𝑃(𝐴∩𝐵)=𝑃(𝐴)×𝑃(𝐵)P(A∩B)=P(A)×P(B)

    2. Conditional Probability Definition: Two events are independent if the occurrence of one event does not change the probability of occurrence of another event. This is expressed as:

       𝑃(𝐴∣𝐵)=𝑃(𝐴)and𝑃(𝐵∣𝐴)=𝑃(𝐵)P(A∣B)=P(A)andP(B∣A)=P(B)

       If these equalities hold, A and B are independent.

    Numerical Problems

    Problem 1: Coin Toss and Die Roll

    Question: You toss a fair coin and roll a fair six-sided die. What is the probability of getting a head and a die roll of 4?

    Solution:

    • 𝑃(Head)=12P(Head)=21​
    • 𝑃(Die roll of 4)=16P(Die roll of 4)=61​
    • Since these events are independent:𝑃(Head and 4)=𝑃(Head)×𝑃(Die roll of 4)=12×16=112P(Head and 4)=P(Head)×P(Die roll of 4)=21​×61​=121​

    Problem 2: Drawing Cards

    Question: If two cards are drawn from a deck of 52 cards, with the first card being replaced before drawing the second, what is the probability of drawing an Ace both times?

    Solution:

    • 𝑃(Ace on first draw)=452=113P(Ace on first draw)=524​=131​
    • Since the card is replaced, the draws are independent:𝑃(Ace on second draw)=452=113P(Ace on second draw)=524​=131​
    • Probability of both events happening:𝑃(Two Aces)=113×113=1169P(Two Aces)=131​×131​=1691​

    Problem 3: Rain and Traffic

    Question: The probability that it will rain on a particular day is 0.30. The probability of encountering traffic regardless of rain is 0.40. Assuming the events are independent, what is the probability it will rain and there will be traffic?

    Solution:

    • 𝑃(Rain)=0.30P(Rain)=0.30
    • 𝑃(Traffic)=0.40P(Traffic)=0.40
    • Since the events are independent:𝑃(Rain and Traffic)=0.30×0.40=0.12P(Rain and Traffic)=0.30×0.40=0.12

    These examples show how independent events work in different contexts. You can see that understanding independence is crucial in determining probabilities in more complex scenarios.

16. 16.Exercise Questions

    Question 1

    Question: If 𝐴A and 𝐵B are two events such that 𝑃(𝐴)=14P(A)=41​, 𝑃(𝐵)=12P(B)=21​, and 𝑃(𝐴∩𝐵)=18P(A∩B)=81​, find 𝑃(not 𝐴 and not 𝐵)P(not A and not B).

    Solution:

    1. We start by finding 𝑃(not 𝐴)P(not A) and 𝑃(not 𝐵)P(not B):

       𝑃(not 𝐴)=1−𝑃(𝐴)=1−14=34P(not A)=1−P(A)=1−41​=43​𝑃(not 𝐵)=1−𝑃(𝐵)=1−12=12P(not B)=1−P(B)=1−21​=21​

    2. We need to find 𝑃(not 𝐴 and not 𝐵)P(not A and not B), which is the same as 𝑃(not (𝐴 or 𝐵))P(not (A or B)) due to De Morgan's Law:

       𝑃(𝐴 or 𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)=14+12−18=58P(A or B)=P(A)+P(B)−P(A∩B)=41​+21​−81​=85​𝑃(not (𝐴 or 𝐵))=1−𝑃(𝐴 or 𝐵)=1−58=38P(not (A or B))=1−P(A or B)=1−85​=83​

    Therefore, 𝑃(not 𝐴 and not 𝐵)=38P(not A and not B)=83​.

    Question 2

    Question: Events 𝐴A and 𝐵B are such that 𝑃(𝐴)=12P(A)=21​, 𝑃(𝐵)=712P(B)=127​, and 𝑃(not 𝐴 or not 𝐵)=14P(not A or not B)=41​. State whether 𝐴A and 𝐵B are independent.

    Solution:

    1. We calculate 𝑃(𝐴 or 𝐵)P(A or B) using 𝑃(not 𝐴 or not 𝐵)P(not A or not B) from De Morgan's Law:

       𝑃(𝐴 or 𝐵)=1−𝑃(not 𝐴 or not 𝐵)=1−14=34P(A or B)=1−P(not A or not B)=1−41​=43​

    2. We then check the independence by calculating 𝑃(𝐴∩𝐵)P(A∩B) using:

       𝑃(𝐴∩𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴 or 𝐵)=12+712−34P(A∩B)=P(A)+P(B)−P(A or B)=21​+127​−43​

       Simplifying the expression:

       𝑃(𝐴∩𝐵)=612+712−912=412=13P(A∩B)=126​+127​−129​=124​=31​

    3. We compare 𝑃(𝐴∩𝐵)=13P(A∩B)=31​ with 𝑃(𝐴)×𝑃(𝐵)=12×712=724P(A)×P(B)=21​×127​=247​. Since 13≠72431​=247​, the events 𝐴A and 𝐵B are not independent.

    By following these steps, we see how the given probabilities influence the independence of events and the outcomes of their combinations.

17. 17.Bayes' Theorem

    Bayes' Theorem is a fundamental concept in probability theory that allows us to update our initial beliefs about the probability of an event based on new evidence. It's named after Thomas Bayes, an 18th-century British statistician and minister. This theorem is especially powerful in fields like medical testing, where it helps in understanding how the probability of a disease changes after obtaining a test result.

    What is Bayes' Theorem?

    Bayes' Theorem calculates the probability of an event based on prior knowledge of conditions that might be related to the event. The formula is:

    𝑃(𝐴∣𝐵)=𝑃(𝐵∣𝐴)×𝑃(𝐴)𝑃(𝐵)P(A∣B)=P(B)P(B∣A)×P(A)​

    Here:

    • 𝑃(𝐴∣𝐵)P(A∣B) is the probability of event 𝐴A given that 𝐵B has occurred.
    • 𝑃(𝐵∣𝐴)P(B∣A) is the probability of event 𝐵B given that 𝐴A has occurred.
    • 𝑃(𝐴)P(A) is the probability of event 𝐴A occurring on its own.
    • 𝑃(𝐵)P(B) is the probability of event 𝐵B occurring on its own.

    Real-World Application

    Consider medical testing, for instance. If a disease affects 1 out of every 1000 people, and a test for the disease is 99% accurate:

    • 𝑃(𝐴)P(A) (disease presence) is 0.001.
    • 𝑃(not 𝐴)P(not A) (disease absence) is 0.999.
    • 𝑃(𝐵∣𝐴)P(B∣A) (test is positive given disease is present) is 0.99.
    • 𝑃(𝐵∣not 𝐴)P(B∣not A) (test is positive given disease is absent) is 0.01.

    Using Bayes' Theorem, we can find out the probability that a person has the disease if they tested positive, adjusting our initial estimate based on the test outcome.

    Careers and Industries

    Bayes' Theorem is used in various fields:

    • Healthcare: Diagnosing diseases from medical test results.
    • Machine Learning: Classifying data based on probabilities.
    • Finance: Assessing risks and probabilities of financial outcomes.
    • Criminal Justice: Evaluating the likelihood of scenarios based on evidence.

    Problem Solving

    Problem 1:
    A disease affects 1 in 1000 people. A test for the disease is 99% accurate. If a person tests positive, what is the probability they actually have the disease?

    Solution:

    1. Calculate 𝑃(𝐴)=0.001P(A)=0.001 and 𝑃(not 𝐴)=0.999P(not A)=0.999.
    2. 𝑃(𝐵∣𝐴)=0.99P(B∣A)=0.99 and 𝑃(𝐵∣not 𝐴)=0.01P(B∣not A)=0.01.
    3. Use Bayes' Theorem: 𝑃(𝐴∣𝐵)=0.99×0.0010.99×0.001+0.01×0.999≈0.0902P(A∣B)=0.99×0.001+0.01×0.9990.99×0.001​≈0.0902

    So, there's about a 9.02% chance they have the disease if they tested positive.

    Problem 2:
    In a survey, 40% of people own a car. If a person owns a car, there is a 70% chance they have a garage. If a person does not own a car, there is a 10% chance they have a garage. What is the probability that a person owns a car given they have a garage?

    Solution:

    1. 𝑃(𝐴)=0.4P(A)=0.4 and 𝑃(not 𝐴)=0.6P(not A)=0.6.
    2. 𝑃(𝐵∣𝐴)=0.7P(B∣A)=0.7 and 𝑃(𝐵∣not 𝐴)=0.1P(B∣not A)=0.1.
    3. Use Bayes' Theorem: 𝑃(𝐴∣𝐵)=0.7×0.40.7×0.4+0.1×0.6≈0.875P(A∣B)=0.7×0.4+0.1×0.60.7×0.4​≈0.875

    There's an 87.5% probability that a person owns a car if they have a garage.

    Problem 3:
    An email filtering system marks emails as spam or not spam. 2% of emails are spam. If an email is spam, there's an 80% chance it gets marked as spam. If an email is not spam, there's a 3% chance it still gets marked as spam. What is the probability an email is spam if it gets marked as spam?

    Solution:

    1. 𝑃(𝐴)=0.02P(A)=0.02 and 𝑃(not 𝐴)=0.98P(not A)=0.98.
    2. 𝑃(𝐵∣𝐴)=0.8P(B∣A)=0.8 and 𝑃(𝐵∣not 𝐴)=0.03P(B∣not A)=0.03.
    3. Use Bayes' Theorem: 𝑃(𝐴∣𝐵)=0.8×0.020.8×0.02+0.03×0.98≈0.34P(A∣B)=0.8×0.02+0.03×0.980.8×0.02​≈0.34

    There's a 34% chance an email is actually spam if it's marked as spam.

    These examples show how Bayes' Theorem helps us make more informed decisions based on probability, refining our initial assumptions with real data. This tool is invaluable in many fields that require analysis under uncertainty

18. 18.Exercise Questions

    Problem:

    A manufacturer has three machine operators A, B, and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time, and C is on the job for 20% of the time. If a defective item is produced, what is the probability that it was produced by A?

    Solution:

    1. Define the Events:

       • Let 𝐷D be the event that a defective item is produced.
       • Let 𝐴A, 𝐵B, and 𝐶C represent the events that the item is produced by operators A, B, and C respectively.

    2. Given Probabilities:

       • 𝑃(𝐴)=0.5P(A)=0.5 (A works 50% of the time)
       • 𝑃(𝐵)=0.3P(B)=0.3 (B works 30% of the time)
       • 𝑃(𝐶)=0.2P(C)=0.2 (C works 20% of the time)
       • 𝑃(𝐷∣𝐴)=0.01P(D∣A)=0.01 (1% defective items from A)
       • 𝑃(𝐷∣𝐵)=0.05P(D∣B)=0.05 (5% defective items from B)
       • 𝑃(𝐷∣𝐶)=0.07P(D∣C)=0.07 (7% defective items from C)

    3. Use Total Probability Theorem to find 𝑃(𝐷)P(D):

       𝑃(𝐷)=𝑃(𝐷∣𝐴)𝑃(𝐴)+𝑃(𝐷∣𝐵)𝑃(𝐵)+𝑃(𝐷∣𝐶)𝑃(𝐶)P(D)=P(D∣A)P(A)+P(D∣B)P(B)+P(D∣C)P(C)𝑃(𝐷)=(0.01)(0.5)+(0.05)(0.3)+(0.07)(0.2)=0.005+0.015+0.014=0.034P(D)=(0.01)(0.5)+(0.05)(0.3)+(0.07)(0.2)=0.005+0.015+0.014=0.034

    4. Use Bayes' Theorem to find 𝑃(𝐴∣𝐷)P(A∣D), the probability that the defective item was produced by A given that a defective item is produced:

       𝑃(𝐴∣𝐷)=𝑃(𝐷∣𝐴)𝑃(𝐴)𝑃(𝐷)P(A∣D)=P(D)P(D∣A)P(A)​𝑃(𝐴∣𝐷)=(0.01)(0.5)0.034≈0.1471P(A∣D)=0.034(0.01)(0.5)​≈0.1471

    So, if a defective item is produced, there is approximately a 14.71% chance that it was produced by operator A.

19. 19.Exercise Questions

    Problem 1:

    There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time, and the third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads. What is the probability that it was the two-headed coin?

    Solution:

    1. Define the Events:

       • Let 𝐴A be the event that the two-headed coin is chosen.
       • Let 𝐵B be the event that the biased coin is chosen.
       • Let 𝐶C be the event that the unbiased coin is chosen.
       • Let 𝐻H be the event that the coin shows heads.

    2. Given Probabilities:

       • 𝑃(𝐴)=13P(A)=31​ (since each coin has an equal probability of being chosen)
       • 𝑃(𝐵)=13P(B)=31​
       • 𝑃(𝐶)=13P(C)=31​
       • 𝑃(𝐻∣𝐴)=1P(H∣A)=1 (two-headed coin will always show heads)
       • 𝑃(𝐻∣𝐵)=0.75P(H∣B)=0.75 (biased coin shows heads 75% of the time)
       • 𝑃(𝐻∣𝐶)=0.5P(H∣C)=0.5 (unbiased coin shows heads 50% of the time)

    3. Use Total Probability Theorem to find 𝑃(𝐻)P(H):

       𝑃(𝐻)=𝑃(𝐻∣𝐴)𝑃(𝐴)+𝑃(𝐻∣𝐵)𝑃(𝐵)+𝑃(𝐻∣𝐶)𝑃(𝐶)P(H)=P(H∣A)P(A)+P(H∣B)P(B)+P(H∣C)P(C)𝑃(𝐻)=(1×13)+(0.75×13)+(0.5×13)=13+0.753+0.53=2.253=0.75P(H)=(1×31​)+(0.75×31​)+(0.5×31​)=31​+30.75​+30.5​=32.25​=0.75

    4. Use Bayes' Theorem to find 𝑃(𝐴∣𝐻)P(A∣H):

       𝑃(𝐴∣𝐻)=𝑃(𝐻∣𝐴)𝑃(𝐴)𝑃(𝐻)P(A∣H)=P(H)P(H∣A)P(A)​𝑃(𝐴∣𝐻)=1×130.75=13÷0.75=13×43=49P(A∣H)=0.751×31​​=31​÷0.75=31​×34​=94​

    The probability that the two-headed coin was chosen, given that the coin shows heads, is 4994​.

    Problem 2:

    An insurance company insured 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. The probability of accidents are 0.01, 0.03, and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

    Solution:

    1. Define the Events:

       • Let 𝑆S be the event that the person is a scooter driver.
       • Let 𝐴A be the event that the person meets with an accident.

    2. Given Probabilities:

       • 𝑃(𝑆)=200012000=16P(S)=120002000​=61​ (total insured is 12000)
       • 𝑃(𝐴∣𝑆)=0.01P(A∣S)=0.01

    3. Use Total Probability Theorem to find 𝑃(𝐴)P(A):

       𝑃(𝐴)=𝑃(𝐴∣𝑆)𝑃(𝑆)+𝑃(𝐴∣𝐶)𝑃(𝐶)+𝑃(𝐴∣𝑇)𝑃(𝑇)P(A)=P(A∣S)P(S)+P(A∣C)P(C)+P(A∣T)P(T)

       Where:

       • 𝑃(𝐶)=400012000=13P(C)=120004000​=31​ (car drivers)
       • 𝑃(𝑇)=600012000=12P(T)=120006000​=21​ (truck drivers)
       • 𝑃(𝐴∣𝐶)=0.03P(A∣C)=0.03
       • 𝑃(𝐴∣𝑇)=0.15P(A∣T)=0.15
       𝑃(𝐴)=(0.01×16)+(0.03×13)+(0.15×12)=0.00167+0.01+0.075=0.08667P(A)=(0.01×61​)+(0.03×31​)+(0.15×21​)=0.00167+0.01+0.075=0.08667

    4. Use Bayes' Theorem to find 𝑃(𝑆∣𝐴)P(S∣A):

       𝑃(𝑆∣𝐴)=𝑃(𝐴∣𝑆)𝑃(𝑆)𝑃(𝐴)P(S∣A)=P(A)P(A∣S)P(S)​𝑃(𝑆∣𝐴)=0.01×160.08667≈0.0193P(S∣A)=0.086670.01×61​​≈0.0193

    The probability that the person is a scooter driver, given that he met with an accident, is approximately 0.0193.