06_Discrete Probability Distributions

The variance of X+Y is equal to the sum of the individual variances: Var(X+Y) = Var(X) + Var(Y)

A random variable is a variable whose actual value is determined by chance operations and is a function that assigns numeric values to different events in a sample space.

The two classes of random variables are:

1. Discrete - can assume only a finite or countably infinite set of numeric values.
2. Continuous - can assume any value within an interval or intervals of real numbers.

Discrete random variable

Continuous random variable

To calculate E(X), use the formula:
E(X) = Σ [x * f(x)]
Where:

• x = number of eggs
• f(x) = density function

Calculating:
E(X) = (3 * 0.35) + (4 * 0.45) + (5 * 0.16) + (6 * 0.04)
E(X) = 1.05 + 1.80 + 0.80 + 0.24 = 3.89

Thus, E(X) = 3.89 eggs.

μ = E(X) = np

Var(X) = np(1 – p)

Discrete random variable

Discrete random variable

Continuous random variable

The probability mass function (PMF) is a mathematical relationship that assigns a probability to each possible value r of a discrete random variable X, defined as:

f(r) = Pr(X = r)

Key properties include:

• f is defined for all real numbers.
• f(r) ≥ 0, since it represents probability.
• f(r) = 0 for most real numbers, as X is discrete and cannot assume most real values.

The probability distribution for the random variable X is as follows:

The probability distribution for the sum of two rolls of a fair 6-sided die (X) is as follows:

The probability that the sum of the two rolls is 10 is P(X=10) = 3/36 = 1/12.

The probability that the sum of the two rolls is 7 or 11 is:

• P(X=7) = 6/36
• P(X=11) = 2/36

Thus, P(X=7 or X=11) = P(X=7) + P(X=11) = 6/36 + 2/36 = 8/36 = 2/9.

X represents the number of patients out of four who are brought under control by the new antihypertensive drug, taking on values 0, 1, 2, 3, or 4.

The probability-mass function indicates the following probabilities for controlling patients:

This shows that the highest probability is for controlling 3 patients (41.1%), followed by controlling 2 patients (26.5%).

The expected value E(X) of a discrete random variable is calculated using the formula:

E(X) = μ = Σ (x₁ * Pr(X = x₁))

where x₁ are the values the random variable assumes with positive probability.

The expected value (E) of a fair 6-sided die can be calculated using the formula:
E(X) = (1/6) * (1 + 2 + 3 + 4 + 5 + 6) = (1/6) * 21 = 3.5

The expected value of the sum of two rolls of a fair 6-sided die can be calculated as follows:
E(X) = E(X1) + E(X2)
Since E(X1) = E(X2) = 3.5,
E(X) = 3.5 + 3.5 = 7

To calculate the expected value (E[X]) for the hypertension-control example, use the formula:

E[X] = Σ (r * Pr(X = r))

Calculating:

E[X] = (0 * 0.008) + (1 * 0.076) + (2 * 0.265) + (3 * 0.411) + (4 * 0.240)
E[X] = 0 + 0.076 + 0.53 + 1.233 + 0.96
E[X] = 2.799

Thus, the expected value is approximately 2.799.

To calculate the expected value (E[X]) for the otitis media example, use the formula:

E[X] = Σ (r * Pr(X = r))

Calculating:

E[X] = (0 * 0.129) + (1 * 0.264) + (2 * 0.271) + (3 * 0.185) + (4 * 0.095) + (5 * 0.039) + (6 * 0.017)
E[X] = 0 + 0.264 + 0.542 + 0.555 + 0.38 + 0.195 + 0.102
E[X] = 2.038

Thus, the expected value is approximately 2.038.

The expected value of a discrete random variable is the 'average' value, also known as the mean or population mean, denoted by μ.

The variance of a discrete random variable, denoted by σ², is also called the population variance and can be calculated using the formula:

Var(X) = σ² = Σ^R_i=1(xi – μ)² Pr (X = xi)

or alternatively as:

σ² = Ε (Χ²) – [E(X)]².

To calculate σX, first find the variance Var(X) using the formula:
Var(X) = E(X^2) - [E(X)]^2

1. Calculate E(X^2):
   E(X^2) = Σ [x^2 * f(x)]
   E(X^2) = (3^2 * 0.35) + (4^2 * 0.45) + (5^2 * 0.16) + (6^2 * 0.04)
   E(X^2) = (9 * 0.35) + (16 * 0.45) + (25 * 0.16) + (36 * 0.04)
   E(X^2) = 3.15 + 7.20 + 4.00 + 1.44 = 15.79

2. Now calculate the variance:
   Var(X) = E(X^2) - [E(X)]^2
   Var(X) = 15.79 - (3.89)^2
   Var(X) = 15.79 - 15.1521 = 0.6379

3. Finally, calculate the standard deviation:
   σX = √Var(X) = √0.6379 ≈ 0.7986

Thus, σX ≈ 0.80 eggs.

The probability is calculated as follows:

P(X≤ 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= 1/36 + 2/36 + 3/36 + 4/36
= 10/36
= 0.28

The cumulative distribution function (CDF) for a discrete random variable X is defined as:

F(x) = P(X ≤ x), for all real x.

The Cumulative Distribution Function (CDF) for the sum of two dice rolled together can be calculated by determining the probability of each possible sum (from 2 to 12) and accumulating these probabilities. The CDF gives the probability that the sum is less than or equal to a certain value. For example:

To find the probability of rolling an 8 or less with two dice, we sum the probabilities of all outcomes that result in sums from 2 to 8. The probabilities are:

Total Probability = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 = 26/36 = 13/18.

To find the probability of rolling a sum between 4 and 10 (inclusive) with two dice, we consider the probabilities of the sums 4, 5, 6, 7, 8, 9, and 10:

Total Probability = 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 = 30/36 = 5/6.

1. A fixed number n of trials are carried out.
2. Each trial can result in one of two outcomes: success or failure.
3. The probability of success p remains constant across trials, with the probability of failure being 1-p.
4. The trials are independent, meaning the outcome of one trial does not affect another.

The probability density function of a binomial random variable with n trials and probability p of success on any trial is given by:

f(x) = ( \frac{n!}{x!(n-x)!} p^x (1-p)^{(n-x)} )

where:

• n = the number of trials or sample size
• p = the probability of success on a trial

The density function for a binomial distribution is given by:

[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} ]

Where:

• ( n ) = total number of trials (5 births)
• ( k ) = number of successes (number of females)
• ( p ) = probability of success (probability of a female being born, typically 0.5)

Thus, the density function for the number of females in families of size 5 is:

[ P(X = k) = \binom{5}{k} (0.5)^k (0.5)^{5-k} = \binom{5}{k} (0.5)^5 ]

To find the probability of at most 3 females in 5 births, we calculate:

[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) ]

Using the binomial probability formula:

[ P(X = k) = \binom{5}{k} (0.5)^5 ]

Calculating each term:

• ( P(X = 0) = \binom{5}{0} (0.5)^5 = 0.03125 )
• ( P(X = 1) = \binom{5}{1} (0.5)^5 = 0.15625 )
• ( P(X = 2) = \binom{5}{2} (0.5)^5 = 0.3125 )
• ( P(X = 3) = \binom{5}{3} (0.5)^5 = 0.3125 )

Thus: [ P(X \leq 3) = 0.03125 + 0.15625 + 0.3125 + 0.3125 = 0.8125 ]

The probability of exactly 3 females in 5 births is calculated using the binomial probability formula:

[ P(X = 3) = \binom{5}{3} (0.5)^5 ]

Calculating:

• ( P(X = 3) = \binom{5}{3} (0.5)^5 = 10 \times (0.5)^5 = 10 \times 0.03125 = 0.3125 )

To find the probability of at least 3 females in 5 births, we calculate:

[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) ]

Using the binomial probability formula:

[ P(X = k) = \binom{5}{k} (0.5)^5 ]

Calculating each term:

• ( P(X = 3) = 0.3125 ) (calculated previously)
• ( P(X = 4) = \binom{5}{4} (0.5)^5 = 5 \times (0.5)^5 = 5 \times 0.03125 = 0.15625 )
• ( P(X = 5) = \binom{5}{5} (0.5)^5 = 1 \times (0.5)^5 = 0.03125 )

Thus: [ P(X \geq 3) = 0.3125 + 0.15625 + 0.03125 = 0.5 ]

E(X+Y) = E(X) + E(Y)

To find the probability of fewer than 5 mice having muscular dystrophy, we can use the binomial distribution formula. Given that the probability of a mouse having muscular dystrophy is 1/4, we calculate the cumulative probability for 0 to 4 mice having the condition.

The probability of exactly 5 mice having muscular dystrophy can be calculated using the binomial distribution formula:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

where n = 20, k = 5, and p = 1/4. This gives the specific probability for this scenario.

To find the probability of fewer than 8 and more than 2 mice having muscular dystrophy, we need to calculate the cumulative probabilities for 3 to 7 mice. This can be done using the binomial distribution formula for each of these values and summing them up.

The Poisson distribution is a probability distribution that describes the number of events occurring in a fixed interval of time or space. It is particularly useful for modeling discrete random variables that represent counts of events. The Poisson distribution is related to the binomial distribution in that it can be used as an approximation when the number of trials is large and the probability of success is small.

A Poisson process gives rise to a discrete random variable that represents the number of occurrences of an event in a continuous interval of time and space, often associated with rare, random events.

Examples of events that can be modeled by a Poisson process include:

1. Number of radioactive decays in a time interval.
2. Number of sedge plants per sampling quadrat.

1. Events occur one at a time: Two or more events do not occur precisely at the same moment or location.

2. Independence of events: The occurrence of an event in a given period of time or space is independent of the occurrence in any previous or later nonoverlapping period.

3. Constant expected number of events: The expected number of events during any one period is the same as during any other period, denoted by μ.

The probability density function is given by:

f(x) = (e^(-μ) * (μ)^x) / x! where x = 0, 1, 2, ...

Here, e is the natural exponential approximately equal to 2.71828.

The expected value E(X) of a Poisson random variable X is equal to μ.

The variance Var(X) of a Poisson random variable X is also equal to μ.

To find the probability of emitting 2 particles in a 1-second interval, we can use the Poisson distribution formula:

P(X=k) = (λ^k * e^(-λ)) / k!

Where:

• λ (lambda) = average rate of occurrence (4 particles per second)
• k = number of occurrences (2 particles)

Substituting the values: P(X=2) = (4^2 * e^(-4)) / 2! = (16 * e^(-4)) / 2 = 8 * e^(-4)

Thus, the probability of emitting 2 particles in a 1-second interval is approximately 0.1465.

Using the Poisson distribution formula, the probability of catching no fish (k=0) is calculated as follows:

P(X = 0) = (λ^k * e^(-λ)) / k!

Where:

• λ = average number of fish caught (2)
• k = number of fish caught (0)
• e = Euler's number (approximately 2.71828)

Thus, P(X = 0) = (2^0 * e^(-2)) / 0! = e^(-2) ≈ 0.1353.

To find the probability of catching fewer than 6 fish, we sum the probabilities of catching 0 to 5 fish using the Poisson distribution:

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Each probability can be calculated using the Poisson formula:

• P(X = k) = (2^k * e^(-2)) / k! for k = 0, 1, 2, 3, 4, 5.

Calculating these gives:

• P(X = 0) ≈ 0.1353
• P(X = 1) ≈ 0.2707
• P(X = 2) ≈ 0.2707
• P(X = 3) ≈ 0.1804
• P(X = 4) ≈ 0.0902
• P(X = 5) ≈ 0.0361

Summing these probabilities gives P(X < 6) ≈ 0.1353 + 0.2707 + 0.2707 + 0.1804 + 0.0902 + 0.0361 ≈ 0.9934.

To find the probability of catching between 3 and 6 fish, we calculate:

P(3 ≤ X ≤ 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

Using the Poisson distribution:

• P(X = 3) ≈ 0.1804
• P(X = 4) ≈ 0.0902
• P(X = 5) ≈ 0.0361
• P(X = 6) = (2^6 * e^(-2)) / 6! ≈ 0.0120

Thus, P(3 ≤ X ≤ 6) ≈ 0.1804 + 0.0902 + 0.0361 + 0.0120 ≈ 0.3187.

To find the probability of at least 1 baby born with the defect, we can use the complement rule:

1. Calculate the probability of no babies being born with the defect:

   • P(no defect) = (1 - p)^n = (1 - 0.0001)^5000

2. The probability of at least 1 baby with the defect is:

   • P(at least 1 defect) = 1 - P(no defect)

To find the probability of no more than 2 babies born with the defect, we can sum the probabilities of having 0, 1, or 2 babies with the defect:

1. Calculate the probabilities using the binomial formula:

   • P(X=k) = (n choose k) * p^k * (1-p)^(n-k)

2. For k = 0, 1, and 2:

   • P(X=0) = (5000 choose 0) * (0.0001)^0 * (0.9999)^5000
   • P(X=1) = (5000 choose 1) * (0.0001)^1 * (0.9999)^4999
   • P(X=2) = (5000 choose 2) * (0.0001)^2 * (0.9999)^4998

3. Sum these probabilities to get P(X ≤ 2).

The conditions for a good approximation are:

1. The binomial parameter n must be large (at least 20).
2. The probability p must be small (no greater than 0.05).
3. The approximation is very good if n ≥ 100 and np ≤ 10.

In a Poisson approximation, the mean μ is equal to np, where n is the number of trials and p is the probability of success in each trial.