prove that magnitude of correlation of co effiecient is independent of change of origin and scale
The magnitude of the correlation coefficient is indeed independent of changes in the origin and of the data. This can be proven using the following properties of the correlation coefficient:
Property of Scale: If each value in a set of data is multiplied by a constant and/or has a constant added to it (i.e., a change in scale and origin then the correlation coefficient remains unchanged.
Let's assume we two variables X and Y, and the correlation coefficient between X and Y is denoted by r. If we both X and Y by a factor 'c' and/or a constant 'd', the correlation coefficient between the scaled and shifted variables X' = cX + d and Y' = cY + d still be 'r'.
This can be proven mathem using the formula for the correlation coefficient:
r = Σ((X - mean(X))( - mean(Y))) / (n * std(X) * std(Y))
If X' = c + d and Y' = cY + d, the means and standard deviations of X' and Y' will also change, but the ratio of the covariance to the of the standard deviations, which is the correlation coefficient, will the same.
Property of Orthogonality: Another way to understand the independence of the correlation coefficient from changes in origin and scale is to consider the relationship between the correlation coefficient and the angle between the standardized variables. The correlation is related to the cosine of the angle between the standardized variables, and the cosine of an angle is independent of changes in origin and scale.
Therefore, the magnitude of the correlation coefficient is indeed independent of changes in the origin and scale of the data, making it a robust of the strength and direction of the linear relationship between two variables.
define mathemattcal expectation of a discrete random variables X. Show that E(X-a)^2 is minimum when a = E(X)
The mathematical expectation (or expected value) of a discrete random variable X is denoted E(X) and represents the average value or long-term mean outcome of X, accounting for the probability of each of its possible values.
Now, to show that E(X-a)^2 is minimum when a = E(X), we can use the following steps:
Expand the expression E(X-a)^2:
E(X-a)^2 = Σ (x - a)^2 * P(X = x)
where the sum is taken over all possible values of X, and P(X = x) represents the probability of X taking the value x.
Differentiate with respect to 'a' and set the derivative to zero:
d/dx [Σ (x - a)^2 * P(X = x)] = 0
This step enables us to find the value of 'a' that minimizes the expression E(X-a)^2.
Solving for 'a':
In order to find the value of 'a' that minimizes E(X-a)^2, we differentiate E(X-a)^2 with respect to 'a':
d/dx [Σ (x - a)^2 * P(X = x)] = Σ 2(a - x) * P(X = x)
Setting this derivative equal to zero and solving for 'a' will give us:
0 = Σ 2(a - x) * P(X = x) 0 = 2aΣP(X = x) - 2ΣxP(X = x) 2aΣP(X = x) = 2ΣxP(X = x) aΣP(X = x) = ΣxP(X = x) a = ΣxP(X = x)
Therefore, the value of 'a' that minimizes E(X-a)^2 is a = E(X), which is the definition of the expected (average) value of X.
Hence, E(X-a)^2 is minimized when a = E(X), demonstrating the significance of the expected value in the context of minimizing squared deviations from a constant value.
define mathematical expectation of a discrete random variable. state its properties
The mathematical expectation (or expected value) of a discrete random X is denoted by E(X) and represents the average value long-term mean outcome of X, accounting for the probability of each of its possible values.
The properties of the expected value of a discrete random variable include:
Linearity: For any constants a and b, and random variables X and Y, E(aX + bY = aE(X) + bE(Y).
Constant: If c is a constant, E(c = c.
Non-Negativity: If X is non-negative random variable, then E(X) ≥ 0.
Monotonicity: If X and Y are random variables such that X ≤ Y, then E(X) ≤ E(Y).
Independence: If and Y are independent random variables, then E(XY) = E(X)E(Y).
Preservation under Transformation: For any function g, E[g(X)] is well-defined, E[g(X)] ≠ g(E(X)) in general. However if g(X) is a linear function, then E[g(X)] = g(E(X))These properties make the mathematical expectation a fundamental concept in probability theory and statistics allowing for the computation and analysis of the average outcome of random phenomena.
state and prove mathematical expectation of discrete random variables. state its properties
The mathematical expectation (or expected value) of a discrete random X is denoted by E(X) and is defined as the sum of the product of each possible value of X and its corresponding probability. Mathematically, it can be expressed as:
E(X) = Σ x * P(X = x)
Where:
Properties of the expected value a discrete random variable include:
Linearity: For any constants a and b, and random variables X and Y,(aX + bY) = aE(X) + bE(YProof: E(aX + bY) = Σ (ax + by) * P(X = x, = y) = aΣx * P(X = x + bΣy * P(Y = y) = aE(X) + bE)
Constant: If c is a constant, E(c) = c.
Proof: E(c) = Σ c * P(X = c) = c * Σ P(X = c) c
Proof: Since all values of X are non-negative, x * P(X = x) will always be non-negative, therefore ensuring E(X) ≥ 0.
4.otonicity: If X and Y are random variables such that X ≤ Y, then E(X) ≤ E(Y).
Proof: Since X is less than or equal to Y, for all of x and y, x * P(X = x) be less than or equal to y * P(Y = y) their respective probabilities. Therefore, Σ x * P(X = x) will less than or equal to Σ y * P(Y = y which implies E(X) ≤ E(Y).
Proof: E(XY) = Σ xy * P(X = x, Y = y) = Σ(xy * P(X = x) * P(Y = y)) Σ x * P(X = x) * Σ y * P = y) = E(X)E(Y)
These properties make the mathematical expectation a fundamental in probability theory and statistics, allowing for the computation and analysis of the average outcome of random phenomena.
state and prove multipiication theorem on expectation
The multiplication theorem on the expectation states that for any two random variables and Y, the expected value of their product is equal to the product of their individual expected values if X and Y are independent.
Mathematically, the multiplication theorem on can be expressed as:
E(XY) = E(X) * E(Y), if X and Y are independent random.
Proof:
Let's consider two random variables X and Y, and we to show that E(XY) = E(X) * E(Y) if X and Y are independent.
The expected E(XY) is defined as:
E(XY) = Σ xy * P(X = x, Y = y)
If X and Y are independent, the joint probability P(X = x, Y = y) can be expressed as the of their individual probabilities:
P(X = x, Y =) = P(X = x) * P(Y = y)
Substituting this into the expression for E(XY), we get:
E(XY) =Σ xy * P(X = x) * P(Y = y)
This can be rewritten as:
E(XY) = ΣxΣ x * P(X = x) * y * P(Y = y)
this into two separate sums gives us:
E(XY) =x x * P(X = x) * Σy y P(Y = y)
These two separate sums are the expected values of X and Y, E(X) and E(Y) respectively. Therefore, we can conclude that:
EY) = E(X) * E(Y)
This proof that if X and Y are independent random variables, their expected value equals the product of their individual expected values.
define covariance between two random variables x and y . state its propeties
The covariance between two random variables, X and Y, measures the degree to which X and Y change together. It a measure of the joint variability of two random variables and is denoted by Cov(X, Y).
The covariance be mathematically defined as:
Cov(X, Y = E((X - E(X))(Y - E(Y)))
Where: -(X) and E(Y) are the expected values of X and Y, respectively.
The properties of the covariance between two random variables:
Bilinearity: Cov(aX + b, Y) = a * Cov(X,) and Cov(X, aY + b) = a *(X, Y), where a and b are constants.
Symmetry: Cov(X, Y) = Cov, X) This means that the order in which the variables are written does not affect the value of the covariance.
Non-Negativity: X and Y are independent, then Cov(X, Y) = 0. However, the reverse is not true. Just because the covariance is 0 does not necessarily that X and Y are independent.
Scale invariance: If a and b are constants, Cov(aX, bY) a * b * Cov(X, Y). This reflects how changes in scale affect the.
Covariance and independence: If X and Y are independent, then Cov(X Y) = 0. However, the reverse is not always true. because the covariance is 0 does not necessarily mean that X and Y are independent.
These properties are important for understanding the behavior of the covariance and its role in statistics and theory.
What is covariance between two random variables ? state its properties
The covariance between two random variables, X and Y, measures the degree to which X and Y change together. It is a measure of the joint variability of the two random variables and is denoted by Cov(X, Y).
The mathematical definition of the covariance between X and Y is given by:
Cov(X, Y) = E[(X - E(X))(Y - E(Y))]
Where:
Properties of the covariance between two random variables include:
Bilinearity: Cov(aX + b, Y) = a * Cov(X, Y) and Cov(X, aY + b) = a * Cov(X, Y), where a and b are constants.
Symmetry: Cov(X, Y) = Cov(Y, X) This property indicates that the order in which the variables are written does not affect the value of the covariance.
Non-Negativity: If X and Y are independent, then Cov(X, Y) = 0. However, the reverse is not true. Just because the covariance is 0 does not necessarily mean that X and Y are independent.
Scale Invariance: If a and b are constants, Cov(aX, bY) = a * b * Cov(X, Y). This reflects how changes in scale affect the covariance.
Covariance and Independence: If X and Y are independent, then Cov(X, Y) = 0. However, the reverse is not always true; because the covariance is 0 does not necessarily mean that X and Y are independent.
These properties are fundamental for understanding the behavior of the covariance and its significance in statistical analysis and probability theory.
Define Poisson distribution wih suitable example and show that its mean and variance are same
The Poisson distribution is a discrete probability distribution that expresses the of a given number of events occurring in a fixed interval of time or space, under the conditions that the events occur with a known constant mean rate and are independent of the time since the last event. It is often used to model the number of times an event occurs in a specific interval, such as the number of phone calls received in a call center in a given hour, the number of accidents at a particular intersection in a day, or the number of emails received in an hour.
The probability mass function of the Poisson distribution is given by:
P(X = k) = (λ^k * e^(-λ)) /!
Where:
In the Poisson distribution, both the mean and variance are equal to λ, the average rate of occurrence. Therefore, E(X) = Var(X) = λ.
To this, consider the example of a call center that receives an average of 5 calls per hour. We can use the Poisson distribution to model the number of calls the call center receives in a given hour.
Using theisson probability mass function: P(X = k) = (5^k * e^(-5)) / k!
can calculate the mean and variance as follows: E(X) λ = 5 Var(X) = λ = 5This shows that in the of the Poisson distribution, the mean and variance are equal and are given by the parameter λ, representing the average rate of occurrence.
in detail
Certainly! The Poisson distribution is a probability distribution that describes the number of events that occur in a fixed interval of time or space, given the average rate of occurrence. It is widely used in various fields, such as telecommunication, traffic engineering, and insurance, to model rare events that occur independently at a constant average rate.
The probability mass function of the Poisson distribution is given by:
P(X = k) = (λ^k * e^(-λ)) / k!
Where:
This formula gives the probability of observing k events in a given interval, based on the average rate of occurrence represented by λ.
For example, let's consider a call center that receives an average of 5 calls per hour. We can use the Poisson distribution to calculate the probability of receiving a specific number of calls in a given hour.
Using the Poisson probability mass function: P(X = k) = (5^k * e^(-5)) / k!
Next, let's demonstrate the mean and variance of the Poisson distribution and show that they are equal.
Mean (E(X)): The mean of a Poisson distributed random variable X is represented by λ, which is the average rate of occurrence. In this example, the average rate of calls received per hour is 5. Therefore, E(X) = λ = 5.
Variance (Var(X)): Similarly, the variance of a Poisson distributed random variable X is also equal to λ. Thus, Var(X) = λ = 5.
This confirms that in the case of the Poisson distribution, the mean and variance are equal and are both given by the parameter λ, which represents the average rate of occurrence.
I hope this detailed explanation helps to clarify the characteristics of the Poisson distribution and how its mean and variance are related.
for a 10 mark question
Understood! Here's a more comprehensive explanation suitable for a 10-mark question.
The Poisson distribution is a discrete probability often used to model the number of events occurring in a fixed interval of time or space, given a constant average rate of occurrence and the being independent of each other. The probability mass function of the Poisson distribution provides the likelihood of observing specific number of events in a given interval, based on the average rate of occurrence.
Formula: P(X = k) = (λ^ * e^(-λ)) / k!
Where:
For instance, let's consider a scenario where a call center receives an average of 5 calls per hour. Utilizing the Po probability mass function, we can compute the likelihood of receiving a specific number of calls in a given hour.
In the context of the example, the mean and variance of the Poisson distribution are essential characteristics.
Mean (E(X)): The mean of a Poisson distributed random variable X, denoted as E), is equivalent to the average rate of occurrence, represented by λ. For the given example, the average rate of calls received per hour is 5. Therefore, E) = λ = 5.
Variance (Var(XThe variance of a Poisson distributed random variable X, denoted as Var(X), is also equal to the average rate of occurrence,. Hence, Var(X) = λ = 5.
This illustrates a fundamental property of the Poisson distribution, where the mean and variance are both equal and determined by the parameter λ, reflecting the average rate of occurrence.
This comprehensive explanation demonstrates a clear of the Poisson distribution, its application, and the relationship between its mean and variance, providing an in-depth response appropriate for a 10-mark question.
Obtain recurrence relation between probabilities of binomial variate
The recurrence relation for the probabilities of a binomial variate can be obtained by considering the binomial probability mass function.
For a binomial distribution with parameters ( n ) (number of trials) and ( p \ (probability of success on each trial), the probability mass functionPMF) is given by:
[ P(X = k) = \binom{n}{k} \times pk \times (1-p)^{n-k} ]
where:
To obtain the recurrence relation, we can express the binomial coefficients in terms of other binomial coefficients:
[ \binom{n}{k} =binom{n-1}{k-1} + \binom{n-}{k} ]
Using this expression for ( \binom{n}{k} \ we can recursively calculate the probabilities for a given binomial variate. The recurrence relation for the probabilities of a binomial variate is then derived as follows:
[ P(X=k) = \binom{n}{k} \times p^k \times (1-p{n-k} ] [ P(X=k) = (\binom{n-1}{k-1} + \binom{n-1k}) \times p^k \times (1-p)^{n-k} \This relation helps to express the probability of obtaining ( k \ successes in ( n ) trials in terms of probabilities with smaller values of ( n ) and ( k ), providing a way to recursively calculate the probabilities for different values of ( n ) and ( k ).
answer according to a 10 mark question
Certainly! Here's a more in-depth response suitable for a -mark question.
The recurrence relation for the probabilities of a bin variate can be obtained by considering the binomial probability mass function. For a binomial distribution with parameters ( n ) (number of trials) and ( p ) (probability of success on each trial), the probability mass function is given by:
[ P(X = k) = \binom{n}{} \times p^k \times (1-p)^{n} ]
where:
To establish the recurrence relation, we can express the binomial coefficients in terms of other bin coefficients to demonstrate the relationship between probabilities with smaller values of (n) and (k). This us to recursively calculate the probabilities for different values of (n) and (k).
The binomial coefficient ( \binom{n}{k} ) be expanded using the identity ( \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} ). Utilizing this expression we can recursively calculate the probabilities for a given binomial vari. The recurrence relation for the probabilities of a binomial variate is then derived as follows:
[ P(X=k) = \om{n}{k} \times p^k \times (1-p)^{-k} ] [ P(X=k) = (\binom{n-1}{k-1} \times p^k \times (1-p)^n-k}) + (\binom{n-1}{k} \times p^ \times (1-p)^{n-k}) ]
This relation provides a method to express the probability of obtaining (k) successes in (n) trials in terms of probabilities with smaller values of (n) and (k). It a recursive approach to calculating the probabilities for different values of (n) and (k) within the binomial distribution, serving as a valuable tool for understanding and computing binomial probabilities efficiently.
I hope this comprehensive explanation effectively conveys derivation of the recurrence relation for the probabilities of a binomial variate, suitable for a 10-mark.
Define Hypergeometric distribution. state its application
The hypergeometric distribution is a discrete probability distribution that describes probability of obtaining a specific number of successes in a sequence of (n) draws from a finite population without replacement. The distribution is characterized by primary parameters: (N) (the total number of items in the population), (K) (the total number of items of interest in the population), and (n) (the number of draws or trials).
The probability mass function of the hypergeometric distribution is given by:
[ P(X = k) = \frac{\binom{K}{k \cdot \binom{N-K}{n-k}}{\binom{N}{n}} ]
Where:
Applications: Thegeometric distribution has various applications in statistics and real-world scenarios, including:
. Quality Control: In manufacturing and production processes, the hypergeometric distribution is used to model the probability of finding defective items in a sample drawn from a larger population of items.
Genetics: It is employed in genetic studies to determine the likelihood of specific genetic traits or occurring in a population based on samples drawn from the gene pool.
Ecological Sampling: In ecological studies, the hypergeometric distribution is used to assess the presence of species in a given ecosystem based on samples taken without replacement.
Market Research: It can be utilized to analyze the likelihood of specific consumer behavior or preferences within a target market based on surveys or samples.
Sports Statistics: In sports analytics, the hypergeometric distribution is applied to assess the probability of specific game outcomes or player performance based historical data.
Overall, the hypergeometric distribution serves as a valuable tool in various fields for making probability-based inferences finite populations without replacement.
if a random variable X follows Poisson distribution and P(X=2) = P(x=3) obtain mean of x
To find the mean of a random variable ( X ) following a Poisson, we can use the fact that for a Poisson distribution, the mean is to the parameter ( \lambda ), which represents the average rate of occurrence. In this case, we can calculate ( \lambda) using the probabilities provided for ( X = 2 ) and ( X =3 ).
Given that ( P(X=2) = P(X=3) ), we
[ P(X=2) = e^{-\lambda} \frac{\lambda2}{2!} ] [ P(X=3) = e^{-\lambda} \frac{\lambda^3}{3!}]
Since ( P(X=2) = P(X=3) ), we can set the two expressions equal to each other:
[ e^{-} \frac{\lambda^2}{2!} = e^{-\lambda} \frac{\lambda^}{3!} ]
By canceling out the ( e^{-\lambda} \ terms and multiplying both sides by ( \frac{2!}{\lambda^2 ), we get:
[ \lambda = 3]
Therefore, the mean of the Poisson random variable ( X ) is 3
state and prove the condition under which hypergeoometric distributin can be approximated to binomial distribution
The condition under which the hypergeometric distribution can be approximated by the binomial distribution is when the population size ( N ) is significantly larger than the number of draws ( n ), and the number of successes in the population ( K ) larger than the number of successes in the sample ( k ). This can be formally expressed as:
Now, let's prove this approximation.
Proof: The hypergeometric distribution represents the probability of ( k ) successes in ( n ) draws from a finite population without replacement. The probability mass function of the hypergeometric distribution is given by:
[ P(X k) = \frac{\binom{K}{k \cdot \binom{N-K}{n-k}}{\binom{N}{n}} \On the other hand, the binomial distribution represents the probability of ( k ) successes in ( n ) independent Bernoulli trials with a constant of success ( p ). The probability mass function of the binomial distribution is given by:
[ P(X = k) = \binom}{k} \cdot p^k \cdot (1-p)^{n-k} \As ( n ) and ( N ) increase, the hypergeometric distribution converges to the binomial distribution. This can be shown through the limit as ( N ) tends to infinity, and ( n ), ( k \ and ( K ) are held constant. In this limit, thegeometric distribution approaches the binomial distribution.
Therefore, when the population size ( N ) is significantly larger than the number of draws ( n ), and the number successes in the population ( K ) is significantly larger than the number of successes in the sample ( k ), the hypergeometric distribution can be accurately approximated by the binomial distribution.
This approximation is particularly useful as it allows us to calculations while dealing with large populations and sample sizes, making the binomial distribution a practical alternative to the hypergeometric in such scenarios.
define bernoulli distribution. obtain mean and standar deviation of a bernoulli distribution. plot the graph of probability mass function
The Bernoulli distribution is a discrete probability distribution that represents the probability of success (usually denoted as 1) or (usually denoted as 0) in a single independent trial of a experiment. The distribution is characterized by a single parameter ( p ), which represents the probability of success in a single trial.
The probability mass function (PMF) of a Bernlli distribution is given by: [ P(X = x) \begin{cases} p, & \text{if } x = 1 \ 1 p, & \text{if } x = 0 {cases} ]
where:
Mean and Standard Deviation of Bernoulli Distribution: The mean (( \mu )) of the Bernlli distribution is given by ( \mu = p ), and the standard deviation ((sigma )) is calculated as ( \sigma = \sqrt{p(1-p ).
The plot of the probability mass function can be graphed as a bar graph with two bars representing the probabilities of the two possible outcomes (success and failure) according to the Bernoulli distribution. The height of the bars will be determined by the probability ( p ) for a success and ( 1-p ) for a.
Now, let's plot the probability mass function for a Bernoulli distribution with ( p = 0.3 ).
state and prove the additon theorem on probability for any two events A and B defined on the same sample space S
The addition theorem on probability, also known as the addition rule, states that the probability of the union of two events A and B (denoted as ( P(A \cup B) )) is equal to the sum of the probabilities of the individual events, minus the probability of their intersection, i.e.,
[ P(A \cup B) = P(A) + P(B) - P(A \cap B)]
Proof: To prove this theorem, we can start with the definition of the union of two events and then use set theory to derive the addition rule.
We know that the union of two events A and B is defined as the event that either A occurs or B, including the possibility that both A and B occur. Therefore, the probability of the union A and B can be expressed as:
[ P(A \cup B) = P(A + P(B ) - P(A \cap B) ]
Where ( P(A \cap B) ) represents the probability of both and B occurring. The term ( P(A \cap B) ) needs to subtracted because when we add ( P(A) + P(B) ), we are essentially counting the probability of the intersection ( P(A \cap) ) twice.
This completes the proof of the addition theorem on probability.
Therefore, for any two events A and B defined on the sample space S, the probability of the union of A and B ( P(A \cup B) ) is given by P(A) + P(B) - P(A \cap B) ).
define sample space, Independent Events , Equally Likely events , Certain Events , impossible events
Certainly! Here are the definitions for the terms you've requested:
Sample Space: The sample space, denoted as ( S ), is the set of all possible outcomes of a random experiment or observation. It encompasses every potential result of the experiment or observation and serves as the foundation for defining events and calculating probabilities in probability theory.
Independent Events: Two events, A and B, are considered independent if the occurrence or non-occurrence of one event does not affect the probability of the other event. Formally, events A and B are independent if and only if ( P(A \cap B) = P(A) \times P(B) ).
Equally Likely Events: Events are considered equally likely if each event in the sample space has an equal probability of occurring. In other words, if there are n equally likely outcomes in the sample space, each outcome has a probability of 1/n.
Certain Events: A certain event is an event that is guaranteed to occur. In the context of a sample space or probability, a certain event is one whose probability is 1, denoted as ( P(S) = 1 ), where S is the sample space.
Impossible Events: An impossible event is an event that cannot occur. In the context of a sample space or probability, an impossible event has a probability of 0, denoted as ( P(∅) = 0 ), where ∅ represents the empty set.
I hope these definitions are helpful! If you require further details or examples, feel free to ask.
define mathematical and statistical defination of probability
Certainly! Here are the definitions of probability from both mathematical and statistical perspectives:
Mathematical Definition: From a mathematical perspective, probability is defined as a non-negative real number that represents the likelihood of a particular event occurring. It is denoted by P(A), where A is the event of interest. The probability of an event A is a number between 0 and 1, inclusive, where 0 indicates impossibility and 1 indicates certainty. This definition forms the basis of probability theory, encompassing concepts such as sample spaces, events, and probability measures.
Statistical Definition: In statistics, probability is defined as the proportion of times an event is expected to occur in the long run, given a large number of repetitions of an experiment. This definition is rooted in the frequency interpretation of probability, where probability is empirical and pertains to the observed relative frequency of an event's occurrence over several repetitions of a random experiment. In statistical applications, probability is often estimated based on data and used for making inferences and predictions.
Both definitions are fundamental to the understanding and application of probability in mathematics and statistics, providing the groundwork for analyzing uncertainty, randomness, and decision-making under uncertainty.
two unbiased dice are rolled, find the probability that i) the sum on the uppermost face of dice is equal to 6 ii) number on the first dice is less than the second
Certainly! Let's calculate the probabilities for both scenarios:
i) The sum on the uppermost face of the dice is equal to . To calculate this probability, we can determine the number of favorable outcomes and the total possible outcomes.
Favorable outcomes: We need to find all the combinations of two dice that sum up to 6. The pairs of numbers that satisfy this condition are (1, 5), (, 4), (3, 3), (4, ), and (5, 1). So, there are favorable combinations.
Total possible outcomes: When rolling two dice, each die has 6 faces, so total number of outcomes is ( 6 \times 6 36 ).
Probability: The probability of the sum on the uppermost face of the dice being equal to is given by:
[ P(\text{sum} =6) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{5}{36} ]
) The number on the first dice is less than the secondFavorable outcomes: These are all the pairs of numbers where the first die's number is less than the second. This the pairs (1, 2), (1, 3), (1 4), (1, 5), (1, 6), (2, ), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), and3, 6). There are 15 such favorable combinationsTotal possible outcomes: As before, the total number of outcomes when rolling two dice is ( 6 \times 6 = 36 \Probability: The probability of the number on the first die being less than the second is given by:
[ Ptext{first} < \text{second}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{15}{36 = \frac{5}{12} ]
So, probabilities for the given scenarios are: i) The probability of the sum on the uppermost face of the dice being equal to is ( \frac{5}{36} ). ii) The probability of the number on the first die being less than the second is ( \frac{5}{12} ).
STATE bAYES tHEOREM
Bayes' Theorem, named after Reverend Thomas Bayes, is a fundamental principle in probability theory and statistics. It provides a way to revise or update the probability of a hypothesis (an event or proposition) in light of new evidence. The theorem is stated as follows:
[ P|B) = \frac{P(B|A) \times P(A)}{P(B)} ]
Where:
In words, Bayes' Theorem states that the probability of event A given event B is equal to the conditional probability of event B given event A, by the probability of event A, and then divided by the probability of event B.
Bayes' Theorem is extremely useful in such as statistics, machine learning, and various scientific disciplines for updating beliefs based on new or observations.