The geometric distribution models the number of trials needed for the first success in independent Bernoulli trials. It's crucial for understanding scenarios with repeated attempts until a desired outcome occurs, like job interviews or manufacturing quality control.
The negative binomial distribution extends this concept, focusing on the number of failures before a specific number of successes. This makes it useful for analyzing more complex situations, such as sales calls or system reliability testing.
Geometric Distribution
Properties of geometric distribution
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Models number of trials needed for first success in independent Bernoulli trials
Bernoulli trials have two possible outcomes (success or failure)
Success probability p p p is constant across trials
Trials are independent
Probability mass function (PMF): P ( X = k ) = ( 1 − p ) k − 1 p P(X = k) = (1 - p)^{k - 1}p P ( X = k ) = ( 1 − p ) k − 1 p , where k k k is number of trials needed for first success
Mean or expected value: E ( X ) = 1 p E(X) = \frac{1}{p} E ( X ) = p 1
Variance: V a r ( X ) = 1 − p p 2 Var(X) = \frac{1 - p}{p^2} Va r ( X ) = p 2 1 − p
Memoryless property: probability of additional trials needed for success is independent of previous failed trials
Geometric distribution probability calculations
Calculate probability of first success on k k k -th trial using PMF: P ( X = k ) = ( 1 − p ) k − 1 p P(X = k) = (1 - p)^{k - 1}p P ( X = k ) = ( 1 − p ) k − 1 p
If p = 0.3 p = 0.3 p = 0.3 , probability of first success on 4th trial is P ( X = 4 ) = ( 1 − 0.3 ) 3 ⋅ 0.3 = 0.1029 P(X = 4) = (1 - 0.3)^{3} \cdot 0.3 = 0.1029 P ( X = 4 ) = ( 1 − 0.3 ) 3 ⋅ 0.3 = 0.1029
Find probability of first success within certain number of trials by summing probabilities for each trial
Probability of first success within 3 trials: P ( X ≤ 3 ) = P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) P(X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3) P ( X ≤ 3 ) = P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 )
Calculate expected number of trials needed for first success using mean: E ( X ) = 1 p E(X) = \frac{1}{p} E ( X ) = p 1
If p = 0.2 p = 0.2 p = 0.2 , expected number of trials is E ( X ) = 1 0.2 = 5 E(X) = \frac{1}{0.2} = 5 E ( X ) = 0.2 1 = 5
Negative Binomial Distribution
Negative binomial vs geometric distributions
Negative binomial models number of failures before r r r -th success in independent Bernoulli trials
r r r is fixed, positive integer representing required successes
Success probability p p p is constant across trials
Trials are independent
Geometric distribution is special case of negative binomial where r = 1 r = 1 r = 1
Negative binomial probability calculations
PMF: P ( X = k ) = ( k + r − 1 r − 1 ) p r ( 1 − p ) k P(X = k) = \binom{k + r - 1}{r - 1}p^r(1 - p)^k P ( X = k ) = ( r − 1 k + r − 1 ) p r ( 1 − p ) k , where k k k is number of failures before r r r -th success
If p = 0.4 p = 0.4 p = 0.4 and we want probability of 3 failures before 2nd success: P ( X = 3 ) = ( 3 + 2 − 1 2 − 1 ) 0. 4 2 ( 1 − 0.4 ) 3 = 0.1296 P(X = 3) = \binom{3 + 2 - 1}{2 - 1}0.4^2(1 - 0.4)^3 = 0.1296 P ( X = 3 ) = ( 2 − 1 3 + 2 − 1 ) 0. 4 2 ( 1 − 0.4 ) 3 = 0.1296
Mean or expected value: E ( X ) = r ( 1 − p ) p E(X) = \frac{r(1 - p)}{p} E ( X ) = p r ( 1 − p )
Variance: V a r ( X ) = r ( 1 − p ) p 2 Var(X) = \frac{r(1 - p)}{p^2} Va r ( X ) = p 2 r ( 1 − p )
Calculate cumulative probabilities by summing individual probabilities for values less than or equal to target value
Applications of geometric and negative binomial distributions
Geometric distribution:
Model number of defective items before finding non-defective item in manufacturing
Determine number of job interviews needed before receiving offer
Analyze number of attempts before successfully completing task (free throw shots in basketball)
Negative binomial distribution:
Model number of unsuccessful attempts before achieving specified successes (sales calls to close certain number of deals)
Analyze failures before system experiences specified successes (password attempts before user successfully logs in certain times)
Determine number of inspections needed to find specified number of defective items in quality control