The binomial and geometric distributions

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Quiz 1




Binomial and Geometric Distributions

Two Special Discrete Distributions

In statistics, there are two discrete distributions commonly used to model situations involving success or failure outcomes: binomial distribution and geometric distribution. Both involve repeated independent trials with a constant probability of success.

Binomial Distribution

The binomial distribution is used to calculate the number of successes in a fixed number of trials. For example, if we roll a die 4 times and want to know how many times we get a six, we can define the random variable R as the number of sixes rolled. Here, R can take values 0, 1, 2, 3, or 4.

Parameters of Binomial Distribution:

  • n: Number of trials (e.g., n = 4)
  • p: Probability of success in each trial (e.g., rolling a six is p = 1/6)
  • q: Probability of failure (q = 1 - p = 5/6)

Probability Formula:

The probability of getting r successes in n trials is given by:

P(R = r) = C(n, r) * p^r * q^(n-r)

Where C(n, r) is the binomial coefficient representing the number of ways to choose r successes from n trials.

Geometric Distribution

The geometric distribution calculates the number of trials needed until the first success occurs. For example, if we roll a die until we get a six, the geometric distribution gives the probability that we get a six on the first roll, second roll, third roll, and so on.

Parameters of Geometric Distribution:

  • p: Probability of success (e.g., rolling a six is p = 1/6)
  • q: Probability of failure (q = 1 - p = 5/6)

Probability Formula:

The probability that the first success occurs on trial number k is given by:

P(X = k) = q^(k-1) * p

Comparison Between Binomial and Geometric Distributions:

Aspect Binomial Geometric
Trials Fixed (n) Until first success
Counts Total successes Trials until success
Formula $$\binom{n}{r}p^r(1-p)^{n-r}$$ $$p(1-p)^{k-1}$$

The binomial and geometric distributions are widely used in practical applications such as risk analysis in investments, product testing, and more. Understanding their differences and applications helps in selecting the appropriate model for specific situations.

Calculating the General Expression for the Number Sequence on the Lily Pad Grid

Problem Description

In a 5x5 grid, a frog can jump only north (N) or east (E). Each lily pad in the grid shows the number of routes the frog can take to reach that pad. To calculate the number of routes to each lily pad, we need to understand that each route consists of a combination of north and east steps.

Calculating the Number of Routes

The number of routes the frog can take to reach a lily pad at position (m, n) (where m is steps north and n is steps east) can be calculated using the combination formula:

Number of Routes = (m+n) choose m = (m+n)! / (m! * n!)

Where:
- m is the number of steps north.
- n is the number of steps east.

Example Calculation

Let's calculate the number of routes to the lily pad at position (2, 2):
- Number of steps north m = 2
- Number of steps east n = 2

The number of routes can be calculated as follows:

Number of Routes = (2+2) choose 2 = 4! / (2! * 2!) = 24 / 4 = 6

Filling the Grid

Here's the 5x5 grid filled with the number of routes to each lily pad:

   1   1   1   1   1
   1   2   3   4   5
   1   3   6   10  15
   1   4   10  20  35
   1   5   15  35  ?

Pattern Discovered

From the grid, we can see a pattern in the numbers. Each number inside the grid is the sum of the two numbers above it, reflecting the properties of Pascal's triangle.

Expression for the nth Term

The numbers on the lily pads form a sequence that can be expressed as:

T(m, n) = (m+n) choose m

Where T(m, n) is the number of routes to the lily pad located at position (m, n).

Expectation and Variance of the Binomial Distribution

The binomial distribution is a probability model used to calculate the number of successes in a fixed number of independent trials, where each trial has two possible outcomes: success or failure. Here's how to calculate the expectation (mean) and variance of the binomial distribution.

Definition and Notation

Let X be a random variable following a binomial distribution with parameters n (number of trials) and p (probability of success on each trial), written as:

X ~ B(n, p)

Expectation (Mean)

The expectation of random variable X following a binomial distribution can be calculated using the formula:

E(X) = n * p

Example:
If we have X ~ B(2, 0.6), then:
E(X) = 2 * 0.6 = 1.2

Variance

The variance of random variable X in a binomial distribution is calculated with the formula:

Var(X) = n * p * (1 - p)

Where q = 1 - p is the probability of failure.

Example:
Using the previous example with n = 2 and p = 0.6:
Var(X) = 2 * 0.6 * (1 - 0.6) = 2 * 0.6 * 0.4 = 0.48

Standard Deviation

The standard deviation is the square root of the variance:

SD(X) = sqrt(Var(X)) = sqrt(n * p * (1 - p))

Example:
From the previous variance calculation:
SD(X) = sqrt(0.48) ≈ 0.69

Conclusion

For a binomial distribution X ~ B(n, p):
- Expectation: E(X) = n * p
- Variance: Var(X) = n * p * (1 - p)
- Standard Deviation: SD(X) = sqrt(n * p * (1 - p))

These formulas are very useful in statistical analysis and help in understanding the behavior of data following a binomial distribution.

Geometric Distribution

The geometric distribution is used to calculate the probability of getting the first success after a number of trials. In this example, we'll look at the situation of rolling a die to get a 6.

Probability of Getting a 6

Let's define:
- p = probability of getting a 6 on one die roll = 1/6
- 1 - p = probability of not getting a 6 = 5/6

Based on this, we can calculate the probability of getting a 6 on a specific trial:

  1. Probability of getting a 6 on the first roll:
    P(X = 1) = p = 1/6
  2. Probability of getting a 6 on the second roll:
    P(X = 2) = (1 - p) · p = (5/6) · (1/6) = 5/36
  3. Probability of getting a 6 on the third roll:
    P(X = 3) = (1 - p)² · p = (5/6)² · (1/6) = 25/216
  4. General formula for the r-th trial:
    P(X = r) = (1 - p)^(r-1) · p
    Where r is the trial number.

Probability Table

Trial (r) Probability P(X=r)
1 p = 1/6
2 (1-p)p = 5/36
3 (1-p)²p = 25/216
4 (1-p)³p = ...
... ...

Properties of Geometric Distribution

  • Independence: Each trial is independent of others.
  • Two Outcomes: Each trial has only two possible outcomes: success (getting a 6) or failure (not getting a 6).
  • Constant Probability: The probability of success remains the same in each trial.

Expectation and Variance

For a geometric distribution with parameter p:
- Expectation (mean): E(X) = 1/p
- Variance: Var(X) = (1 - p)/p²

Application Example

Suppose in a country, 18% of adults wear contact lenses. If we want to find the probability that the first adult who wears contact lenses is one of the first interviewed, we can use the geometric distribution formula.

Example Problem:
a. Probability that the first person wearing contact lenses is one of the first 15 interviewed:
P(X ≤ 15) = 1 - P(X > 15) = 1 - (0.82^15)
b. Probability that the first person wearing contact lenses is not one of the first nine interviewed:
P(X > 9) = (0.82^9)

Mode of the Geometric Distribution

The mode is the most likely value to occur in a probability distribution. In the geometric distribution, the mode is always 1. This means that the highest probability of getting the first success occurs on the first trial.

Why is the Mode Always 1?

The geometric distribution has the property that the probability of success on the r-th trial is calculated by the formula:
P(X = r) = p · (1-p)^(r-1)
Where:
- p: probability of success on each trial
- 1-p: probability of failure

From this formula, we can see that:
- For r = 1, the probability is P(X = 1) = p, which is the highest value.
- For r > 1, the probability will decrease because the factor (1-p)^(r-1) continues to decrease.

Probability Pattern in Geometric Distribution

The probabilities in a geometric distribution form a geometric sequence with the first term p and common ratio 1-p. Since this ratio is always less than 1, the probability values will continue to decrease as the value of r increases.

For example:
If X ~ Geo(0.2), then:
- P(X = 1) = 0.2
- P(X = 2) = 0.2 · (0.8) = 0.16
- P(X = 3) = 0.2 · (0.8)² = 0.128

From this, we can see that the highest probability is on the first trial (r = 1).

The mode of all geometric distributions is 1, because the first trial always has the highest probability of producing the first success. This is due to the basic nature of the geometric distribution, where the probability of success decreases exponentially with the increasing number of trials before success occurs.

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