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Learn how to calculate probabilities using the binomial and hypergeometric distributions with real-life examples. Find probabilities, expected values, and standard deviations.
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Math 3680 Lecture #5 Important Discrete Distributions
Example:A student randomly guesses at three questions. Each question has five possible answers, only once of which is correct. Find the probability that she gets 0, 1, 2 or 3 correct. This is the same problem as the previous one; we will now solve it by means of the binomial formula.
Example:Recall that if X ~ Binomial(3, 0.2), P(X = 0) = 0.512 P(X = 1) = 0.384 P(X = 2) = 0.096 P(X = 3) = 0.008 Compute E(X) and SD(X).
MOMENTS OF Binomial(n, p) DISTRIBUTION E(X) = np SD(X) = Var(X) = Try this for the Binomial(3, 0.2) distribution. Do these formulas make intuitive sense?
Example:A die is rolled 30 times. Let X denote the number of aces that appear. A) Find P(X = 3). B) Find E(X) and SD(X).
Example:Three draws are made with replacement from a box containing 6 tickets: • two labeled “1”, • one each labeled, “2”, “3”, “4” and “5”. Find the probability of getting two “1”s.
Example:Three draws are made without replacement from a box containing 6 tickets: • two labeled “1”, • one each labeled, “2”, “3”, “4” and “5”. Find the probability of getting two “1”s.
P(S1S2S3) = P(S1) P(S2 | S1) P(S3 | S1 ∩ S2) = P(S1S2F3) = P(S1) P(S2 | S1) P(F3 | S1 ∩ S2) = P(S1F2S3) = P(S1) P(F2 | S1) P(S3 | S1 ∩ F2) = P(S1F2F3) = P(S1) P(F2 | S1) P(F3 | S1 ∩ F2) = P(F1S2S3) = P(F1) P(S2 | F1) P(S3 | F1 ∩ S2) = P(F1S2F3) = P(F1) P(S2 | F1) P(F3 | S1 ∩ F2) = P(F1F2S3) = P(F1) P(F2 | F1) P(S3 | F1 ∩ F2) = P(F1F2F3) = P(F1) P(F2 | F1) P(F3 | F1 ∩ F2) =
The Hypergeometric Distribution Suppose that n draws are made without replacement from a finite population of size N which contains G “good” objects and B = N - G “bad” objects. Let X denote the number of good objects drawn. Then where b = n - g.
Example:Three draws are made without replacement from a box containing 6 tickets; two of which are labeled “1”, and one each labeled, “2”, “3”, “4” and “5”. Find the probability of getting two “1”’s.
MOMENTS OF HYPERGEOMETRIC(N, G, n) DISTRIBUTION E(X) = np (where p = G / N) SD(X) = Var(X) =
REDUCTION FACTOR The term is called the Small Population Reduction Factor. It always appears when we draw without replacement. If the population is large (N> 20 n) , then the reduction factor can generally be ignored (why?).
Example: Thirteen cards are dealt from a well-shuffled deck. Let X denote the number of hearts that appear. A) Find P(X = 3). B) Find E(X) and SD(X).
Example. A lonely bachelor decides to play the field, deciding that a lifetime of watching “Leave It To Beaver” reruns doesn’t sound all that pleasant. On 250 consecutive days, he calls a different woman for a date. Unfortunately, through the school of hard knocks, he knows that the probability that a given woman will accept his gracious invitation is only 1%. What is the chance that he will land three dates?