Chapt. 6: Probability Distributions cont.

1. Chapt. 6: Probability Distributionscont.  Random Variables  Probability Distributions The Binomial Distribution The Hypergeometric Distribution • The Poisson Distribution • The Multinomial Distribution • The Mean of a Probability Distribution • The Standard Deviation of a Probability Distribution • Chebyshev’s Theorem

2. Hypergeometric Distribution If the trials are not independent, we cannot apply the binomial probability distribution to find the probability of x successes in n trials. If the trials are dependent we replace the binomial probability distribution by the hypergeometric probability distribution

3. Dependent events If n objects are chosen at random (without replacement) from a set consisting of a objects of type A and b objects of type B Then the probability of xof typeAandn - xof typeB is f(x) =   ·   a x b n - x aCx · bCn - x for x = 0,1,2, … n a+bCn   a + b n Note x  a n - x  b Hypergeometric distribution Rewriting

4. Another way to say this previous formula a + b a b N = total number of elements in the population s = number of successes in the population N- s = number of failures in the population n = number of trials (sample size) x = number of successes in n trial n - x = number of failures in n trials The probability of x successes in n trials is given by n x n - x P(x) =   ·   N - s n - x s x (  N n

5. Example: Airmail stamps A secretary is supposed to send 6 of 15 letters by airmail, but she gets the all mixed up and randomly puts airmail stamps on 6 of the letters. What is the probability that only 3 of the letter that should go by airmail get an airmail stamp ? a = 6, b = 9, n = 6, x = 3 Sampling without replacement ? YES 6 9 3 3     20 · 84 = 5005 f(3) =  0.336   15 6

6. Example: Red Heads Among your 16 friends there are 5 with red hair. If 3 of them are randomly picked find the separate probabilities that these 3 will include 0,1,2 or 3 red heads. Solution:Sampling without replacement ! a = 5 b = 11 n = 3 x = 0,1,2,3 5 11 0 3     1 · 165 f(0) = = ≈ 0.295 560   16 3

7. 5 11 1 2     5 · 55 f(1) = = ≈ 0.491 560   16 3 5 11 2 1     10 · 11 f(2) = = ≈ 0.196 560   16 3 5 11 3 0      10 · 1 f(3) = = ≈ 0.018 560   16 3 Histogram please

8. Histogram : Red Heads f(0) = 0.295 , f(1) = 0.491 , f(2) = 0.196 , f(3) = 0.018 0.491 0.295 0.196 0.018 0 1 2 3

9. Typical card question • If 5 cards are dealt , without replacement, form a standard deck of 52 cards. What is the probability that the 5 cards include exactly 2 spades ? A = spades, B = non-spades a = 13 b = 39 n = 5 x = 2 ( ) ( ) Solution 13 2 39 3 78 · 9139 f(2) = = ≈ 0.274 ( ) 52 5 2,598,960

10. Chapt. 6: Probability Distributions , cont.  Random Variables  Probability Distributions The Binomial Distribution The Hypergeometric Distribution The Poisson Distribution • The Multinomial Distribution • The Mean of a Probability Distribution • The Standard Deviation of a Probability Distribution • Chebyshev’s Theorem

11. Poisson Distribution • The Poisson probability distribution is an approximation to the binomial distribution when n is large and p is small typicallyn ≥ 100andnp < 10 np = average number of successes Example: A washing machine in a laundromat breaks down an average of 3 times a month. Find the probability of exactly 2 breakdown during the next month. breakdown = “success” or “occurrence” in Poisson distribution terminology

12. Special Poisson formula Probability of x successes (occurrences) in n trials f(x) = e-np (np)x x = 0,1,2,3…. x ! n ≥ 100 , np < 10 e = 2.71828….., natural logarithms ! x is a discrete random variable occurrences are random and independent