Binomial Distribution

Binomial Distribution
(discrete random variable)
• Calculating probabilities
– determine sample space
– # events and whether equal probable
– determine the events in the class
– take ratio if equal probable
• otherwise sum the probabilities
• roulette example
– 37 equally spaced slots (0-36)
– 18 red, 18 black, 1 green
– probability of even? divisible by 3? even and red?

Sequences of Events
• A particular sequence of samples (e.g., drawing
marbles)
– joint event across samples
– the sequence can itself be viewed as an elementary event
• Counting rule for N trials of K possible events (sampling
w/ replacement)
– K N possible sequences if K are mutually exclusive and
exhaustive
– Sequences with different K on different trials
• multiply the appropriate K of each trial
• e.g., flipping a coin then rolling a die
• Permutations (sampling w/o replacement)
– ordering N distinct objects
– the number of ordering is N!, where 0!=1
– e.g., 10 students in 10 chairs = 3,628,800
• drawing 2 red then 3 blue marbles?

• Ordered Combinations
– Selecting r objects from N total objects (r ≤ N) when you care
about the order of the r objects (but don’t care about the N-r
remaining objects)
• N! / (N-r)!
• 40 lottery tickets, and 3 winners in order
– # of possible results = ?
• n P r on many calculators
• Unordered Combinations
– “Choosing” r objects from N total objects when you no longer
care about the order in which they’re chosen
⎛N⎞ ⎛N⎞ N !
⎜ ⎟=
⎝ r ⎠ r!( N − r)!
• ⎜ ⎟
⎝ r ⎠is
sometimes called the binomial coefficient
• could be written as nCr • same regardless of which you choose
⎛N⎞ ⎛ N ⎞
⎜ ⎟= ⎜ ⎟
⎝ r ⎠ ⎝N − r⎠

Poker Hands
• how many possible hands?
⎛52⎞ 52!
⎜ ⎟=
= 2,598,960
⎝ 5 ⎠ 5!47!
– roughly 4 in 10 million
• one pair, with 3 different remaining cards
– number of ways to have a pair
– number of ways to have the remaining cards
⎛4⎞ 3 ⎛12⎞ 13⎜ ⎟4⎜ ⎟
2 3
p(one
pair) =
⎝ ⎠ ⎝ ⎠
≈.42
⎛52⎞ ⎜ ⎟
⎝ 5 ⎠
• full house (.0014)? flush? (.00198) nothing?
⎛4⎞ 13⎜ ⎟
⎝2⎠ ⎛12⎞ (4)(4)(4) ⎜ ⎟
⎝ 3 ⎠

Bernoulli Trials
• For Bernoulli process, just two event classes (e.g., coin
flip)
– p = success
– q = failure = 1-p
– stationary versus non-stationary
– not all orders are equally probable (depends on p)
– for independent stationary, the probability of a particular
(ordered) sequence of N trials
• prqN-r , where r is number of successes
• same probability for all orders giving with this number of successes
– often we want to know the probability of r successes, regardless
of order
⎛N⎞ ⎜ ⎟ pq
⎝ r ⎠
• multiply by the number of orders r N−r

Binomial Distribution
• Binomial sampling
– typically, N and p are known, and want probability for different
r successes
• By considering all possible numbers of success (0-N),
we can construct the binomial distribution
– parameters, N and p
• the mathematical rule defines a family of particular distributions
⎛N⎞ r N−r px ( = rN ; , p) = ⎜ ⎟ pq , for r= 0,1, 2,..., N
⎝ r ⎠
– examples with 5 coin flips (p=.5) and 5 female college students
(p=.1 married)

Probabilities of Intervals for Binomial
• for p = .30, N = 10, find p(1 ≤ x ≤ 7)
– same as 1 - p(x = 0, or 8 ≤ x)
– could also list the cumulative
– often we express the binomial in terms of a proportion P = r/N
– appendix F lists binomials (p. 1008), factorials (p. 1028), and
coefficients (1029)

Subliminal Perception
• Guess as to which quadrant a subliminal
point of light is presented
– chance performance is p=.25
– out of 10 trials, 7 are correct
• p(x=7; p=.25, N=10) = .0031
– but even p(x=2) is only .28
• what we want is guessing as compared to having
some information (i.e., x ≥ 7)
– p(x ≥ 7; p=.25, N=10) = .0035
» this is the prob. of “rejecting null hypothesis”
• how would a Bayesian deal with this problem?

Sign Test for 2 Matched Groups
• Use binomial to test for a difference between two matched groups
– e.g., paired/yoked subjects, pre-post test, etc.
• in general is X A (type A) different than X B (type B)?
– for each pair, take the difference x A –x B
– positive differences are +, negative are -, and flip coin for 0 differences (or simply
remove ties and reduce N)
– After training 20 people, 15 improved
• What’s the probability that this would be true if there was no change on
average?
– p(15 or more) = .02
• Can we conclude that the training was effective?
• What if instead we observed improvement in only 5?
– one-tailed versus two-tailed
• Other distributions related to binomial
– geometric: number N needed for r successes
– negative binomial: number of failures before rth success
– Poisson: approximation to binomial with large N and small p
– multinomial: more than 2 outcomes
– hypergeometric: multinomial sampling without replacement