Geometric and Negative Binomial Probability Distributions

Geometric Probability Distribution

• Computes the probability of conducting $x$ trials before observing the first success.

• The random variable $X$ represents the total number of trials required to obtain $1$ success.

• Key Assumptions and Characteristics:

  • The trials are identical and independent of one another.
  • The probability of success ($p$) remains constant across all trials (identical to the Binomial distribution assumption).

• Probability Mass Function (PMF):

  • Formula: $P(X = x) = q^{(x-1)} \times p$
  • Equivalent Form: $P(X = x) = (1 - p)^{(x-1)} \times p$
  • Domain / Support: $x \in \{1, 2, 3, \dots\}$ where $0 < p < 1$
  • Variables:
  • $x$: Total number of trials needed to obtain a success
  • $p$: Probability of success on any individual trial
  • $q$: Probability of failure on any individual trial ($q = 1 - p$)

• Expectation and Variance:

  • Expected Value: $E(X) = \frac{1}{p}$
  • Variance: $\text{Var}(X) = \frac{q}{p^2}$

• Comparison with Binomial Distribution:

  • Assumed context: Let $P(\text{female}) = 0.49$ and $P(\text{male}) = 0.51$
  • Binomial Question Framing: "What is the probability that out of the next $5$ children of a heterosexual couple, $1$ will be female?" (Fixed number of trials $n = 5$, counting total successes).
  • Geometric Question Framing: "What is the probability that a heterosexual couple will have to conceive $5$ children to have $1$ daughter?" (Fixed number of target successes $r = 1$, counting total trials $x = 5$).
  • Geometric Calculation: $P(X = 5) = (0.51)^{5-1} \times 0.49 = (0.51)^4 \times 0.49$

• Worked Examples:

  • Example 16: Rolling a Fair Die
  • Scenario: A fair six-sided die is rolled repeatedly ($p = \frac{1}{6}$, $q = \frac{5}{6}$).
  • Case 1: Probability of getting a $4$ on the $5\text{th}$ roll.
    • Parameters: $x = 5$ attempts, $p = \frac{1}{6}$, $q = \frac{5}{6}$
    • Calculation: $P(X = 5) = \left(\frac{5}{6}\right)^{5-1} \times \left(\frac{1}{6}\right)^1 = \left(\frac{5}{6}\right)^4 \times \left(\frac{1}{6}\right)^1 = 0.0804$
  • Case 2: Probability of having to roll the die $3$ times to finally get a $5$
    • Parameters: $x = 3$ attempts, $p = \frac{1}{6}$, $q = \frac{5}{6}$
    • Calculation: $P(X = 3) = \left(\frac{5}{6}\right)^{3-1} \times \frac{1}{6} = \left(\frac{5}{6}\right)^2 \times \frac{1}{6}$
  • Example 17: Capturing Sea Creatures
  • Scenario: $10\%$ ($0.10$) of creatures in a certain sea are venomous. The goal is to capture $1$ venomous creature.
  • Parameters: Success probability $p = 0.10$, failure probability $q = 0.90$
  • Case 1: Probability of having to capture and test $4$ creatures just to get $1$ venomous creature.
    • Parameters: $x = 4$, $p = 0.10$, $q = 0.90$
    • Calculation: $P(X = 4) = (0.90)^{4-1} \times 0.10 = (0.90)^3 \times 0.10 = 0.0729$
  • Case 2: Probability of having to capture and test $6$ creatures just to get $1$ venomous creature.
    • Parameters: $x = 6$, $p = 0.10$, $q = 0.90$
    • Calculation: $P(X = 6) = (0.90)^{6-1} \times 0.10 = (0.90)^5 \times 0.10 = 0.059049$

Negative Binomial Probability Distribution

• Computes the probability of conducting $x$ trials before achieving $r$ successes.

• Generalizes the Geometric distribution to scenarios requiring more than $1$ success ($r$ successes).

• Key Assumptions and Characteristics:

  • Shares identical characteristics with the Geometric distribution (independent and identical trials with constant probability $p$).
  • Evaluates the total number of trials $x$ required to attain a specific target number of successes $r$

• Probability Mass Function (PMF):

  • Combination Notation: $P(X = x) = \binom{x-1}{r-1} \times q^{(x-r)} \times p^r$
  • Alternative Form: $P(X = x) = [_{x-1}C_{r-1}] \times [q^{(x-r)}] \times [p^r]$
  • Variables:
  • $x$: Total number of trials needed to obtain the $r$ successes
  • $r$: Target number of successes required
  • $p$: Probability of success on any individual trial
  • $q$: Probability of failure on any individual trial ($q = 1 - p$)

• Reduction to Geometric Distribution:

  • If the target number of successes is $r = 1$, the Negative Binomial distribution reduces directly to the Geometric distribution:
  • Substitution: $P(X = x) = \binom{x-1}{1-1} \times q^{(x-1)} \times p^1$
  • Since $\binom{x-1}{0} = 1$:
  • $P(X = x) = 1 \times q^{(x-1)} \times p = q^{(x-1)} \times p$

• Expectation and Variance:

  • Expected Value: $E(X) = \frac{r}{p}$
  • Variance: $\text{Var}(X) = \frac{r \times q}{p^2}$

• Comparison with Binomial Distribution:

  • Assumed context: Let $P(\text{female}) = 0.49$ and $P(\text{male}) = 0.51$
  • Negative Binomial Question Framing: "What is the probability that a heterosexual couple will have to conceive $5$ children to have $2$ daughters?"
  • Parameters: $x = 5$, $r = 2$, $p = 0.49$, $q = 0.51$
  • Formula Application: $P(X = 5) = \binom{5-1}{2-1} \times (0.51)^{(5-2)} \times (0.49)^2 = \binom{4}{1} \times (0.51)^3 \times (0.49)^2$

• Worked Examples:

  • Example 19: Rolling a Fair Die
  • Scenario: Rolling a fair six-sided die repeatedly ($p = \frac{1}{6}$, $q = \frac{5}{6}$).
  • Case 1: Probability of getting two $4$\text{s} by the time the $5\text{th}$ roll is finished.
    • Parameters: $x = 5$, $r = 2$, $p = \frac{1}{6}$, $q = \frac{5}{6}$
    • Calculation: $P(X = 5) = \binom{5-1}{2-1} \times \left(\frac{5}{6}\right)^{5-2} \times \left(\frac{1}{6}\right)^2 = \binom{4}{1} \times \left(\frac{5}{6}\right)^3 \times \left(\frac{1}{6}\right)^2$
  • Case 2: Probability of having to roll the die just $3$ times to get two $5$\text{s}.
    • Parameters: $x = 3$, $r = 2$, $p = \frac{1}{6}$, $q = \frac{5}{6}$
    • Calculation: P(X = 3) = \binom{3-1}{2-1} \t\times \left(\frac{5}{6}\right)^{3-2} \times \left(\frac{1}{6}\right)^2 = \binom{2}{1} \times \left(\frac{5}{6}\right)^1 \times \left(\frac{1}{6}\right)^2
  • Example 20: Capturing Sea Creatures
  • Scenario: $10\%$ ($0.10$) of sea creatures are venomous. The lab experiment requires $3$ venomous creatures ($r = 3$).
  • Case 1: Probability of having to capture and test $4$ creatures to get $3$ venomous creatures.
    • Parameters: $x = 4$, $r = 3$, $p = 0.10$, $q = 0.90$
    • Calculation: $P(X = 4) = \binom{4-1}{3-1} \times (0.90)^{(4-3)} \times (0.10)^3 = \binom{3}{2} \times (0.90)^1 \times (0.10)^3$
  • Binomial Counterpart Framing:
    • Question: "What is the probability that out of $4$ creatures tested, $3$ are venomous?" (In Binomial, the sample size $n = 4$ is fixed prior to testing, whereas in Negative Binomial, the required successes $r = 3$ are fixed).
  • Case 2: Probability of having to capture and test $6$ creatures to get $3$ non-venomous creatures.
    • Parameters: Here success is defined as capturing a non-venomous creature, so $p = 0.90$ and $q = 0.10$. Total trials $x = 6$, required successes $r = 3$
    • Calculation: $P(X = 6) = \binom{6-1}{3-1} \times (0.10)^{(6-3)} \times (0.90)^3 = \binom{5}{2} \times (0.10)^3 \times (0.90)^3$
  • Case 3: Probability that the $5\text{th}$ creature captured is the $2\text{nd}$ (and final needed) venomous creature.
    • Parameters: $x = 5$, $r = 2$, $p = 0.10$, $q = 0.90$
    • Calculation: $P(X = 5) = \binom{5-1}{2-1} \times (0.90)^{(5-2)} \times (0.10)^2 = \binom{4}{1} \times (0.90)^3 \times (0.10)^2$
  • Example 21: Sex of an Unborn Child
  • Applies Negative Binomial principles to analyze probabilities regarding child birth order and gender outcomes.