Question

If X∼Geometric(p),n≥1;0

Answer 1

The probability mass function (PMF) of a geometric distribution is given by
`\( P(X = k) = (1 - p)^(k-1) p \)`, where ( p ) is the probability of success on each trial. Therefore, the expression ( P(X > n) ) can be written as
`\( (1 - p)^n \).`

Step-by-step explanation:

In a geometric distribution with probability of success ( p ), the PMF ( P(X = k) ) represents the probability of having the first success on the ( k )-th trial. The complementary event ( P(X > n) ) is the probability of having the first success occur after the first ( n ) trials. This can be expressed as
`\( 1 - P(X \leq n) \), where \( P(X \leq n) \)` is the cumulative distribution function (CDF) up to ( n ).

The CDF is the sum of the PMF up to ( n ). Therefore, ( P(X > n) ) is the complement of this sum, representing the probability of the first success occurring after ( n ) trials. Mathematically, this is
`\( 1 - \sum_(k=1)^(n) P(X = k) \).`Substituting the PMF into this expression, we get
`\( 1 - \sum_(k=1)^(n) (1 - p)^(k-1) p \).`Simplifying further, this expression equals
`\( (1 - p)^n \)`.

So, the final answer is
`\( (1 - p)^n \)`, which represents the probability of the first success occurring after the first ( n ) trials in a geometric distribution.

Answer 2

The probability that X is at least n, denoted as P(X ≥ n), for a geometric distribution with parameter p, is (1 - p)^(n-1).

Step-by-step explanation:

The geometric distribution models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials, each with probability of success p. For a given random variable X following a geometric distribution, the probability of X being at least n (P(X ≥ n)) can be computed as the complement of the probability of X being less than n.

The probability of X being less than n is the cumulative probability of not achieving success in the first n-1 trials, which is (1 - p)^(n-1). Therefore, to find the probability of X being at least n, we subtract this cumulative probability from 1, giving us P(X ≥ n) = 1 - (1 - p)^(n-1).

For instance, if the probability of success in a single trial is 0.3 and we want to find the probability that the first success occurs on or after the 5th trial, the probability P(X ≥ 5) would be 1 - (1 - 0.3)^4, which simplifies to 1 - 0.7^4, resulting in the probability that X is at least 5.

This formula is derived from understanding that in a geometric distribution, the probability decreases exponentially with each additional trial until the first success occurs, making P(X ≥ n) easily calculable using the complement of the cumulative probability of failure over n-1 trials.