Geometric Distribution

• Let a sequence of Bernoulli trials be performed, each with constant probability \(p\) of success and \(q = 1 – p\) of failure.
• Trials are independent, and we continue performing them until the first success occurs.
• Let \(X\) be the number of failures preceding the first success.
• Then \(X\) can take values \(0, 1, 2, 3, \ldots\)

Probability of Failures before First Success

• \( P(X = 0) = P(\text{zero failure preceding the first success}) = P(S) = p \)
• \( P(X = 1) = P(\text{one failure preceding the first success}) = P(F \cap S) = qp \)
• \( P(X = 2) = P(\text{two failures preceding the first success}) = P(F \cap F \cap S) = q^2p \)
• And so on…

General Form

• Therefore, in general, the probability of \(x\) failures preceding the first success is:
  \[
  P(X = x) = q^x p, \quad x = 0, 1, 2, 3, \ldots
  \]
• The respective probabilities \(p, qp, q^2p, q^3p, \ldots\) form a geometric progression with common ratio \(q\).
• Hence, this probability distribution is known as the Geometric Distribution.

Definition

• A random variable \(X\) follows geometric distribution if:
  \[
  P(X = x) =
  \begin{cases}
  q^x p, & x = 0, 1, 2, \ldots \\
  0, & \text{otherwise.}
  \end{cases}
  \]
• The sum of all probabilities is:
  \[
  \sum_{x=0}^{\infty} q^x p = p(1 + q + q^2 + q^3 + \ldots) = 1.
  \]

Example: Geometric Distribution – Die Problem

• An unbiased die is cast until 6 appears.
• Find the probability that it must be cast more than five times.

Step 1: Define Success and Failure

• Probability of success (getting 6): \( p = \frac{1}{6} \)
• Probability of failure: \( q = 1 – p = \frac{5}{6} \)

Step 2: Define Random Variable

• Let \(X =\) number of failures before first success.
• Then:
  \[
  P(X = x) = q^x p, \quad x = 0, 1, 2, \dots
  \]

Step 3: Desired Probability

• We want \(P(\text{Die is cast more than 5 times}) = P(X \ge 5)\)

Step 4: Express as a Series

• \[
  P(X \ge 5) = P(X = 5) + P(X = 6) + \dots
  \]
• Substitute:
  \[
  = \left(\frac{5}{6}\right)^5 \left(\frac{1}{6}\right)
  \left[1 + \frac{5}{6} + \left(\frac{5}{6}\right)^2 + \dots \right]
  \]

Step 5: Simplify Using Geometric Series

• Infinite geometric series:
  \[
  1 + r + r^2 + \dots = \frac{1}{1 – r}, \quad |r| < 1
  \]
• Here \(r = \frac{5}{6}\).
• Hence:
  \[
  = \left(\frac{5}{6}\right)^5 \left(\frac{1}{6}\right)
  \left(\frac{1}{1 – \frac{5}{6}}\right)
  \]

Step 6: Final Simplification

• Simplifying:
  \[
  = \left(\frac{5}{6}\right)^5
  \]
• Hence, the probability that the die must be cast more than five times is:
  \[
  \boxed{\left(\frac{5}{6}\right)^5}
  \]

Mean of the Geometric Distribution

• Mean:
  \[
  E(X) = \sum_{x=0}^{\infty} x q^x p = p \sum_{x=1}^{\infty} x q^{x-1}
  \]

Final Result

• \[
  E(X) = \frac{q}{p}, \quad V(X) = \frac{q}{p^2}
  \]

Remark 1: Geometric Distribution

Example 2: Comment on the Following

Solution

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