Math 3680 Lecture #5 Important Discrete Distributions.

Math 3680

Lecture #5

Important DiscreteDistributions

The Binomial Distribution

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, ) DISTRIBUTION

E(X) = n

SD(X) =

Var(X) =

Try this for the Binomial(3, 0.2) distribution.

Do these formulas make intuitive sense?

)1( n

)1( n

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.

1 2 3

0.1

0.2

0.3

0.4

The Hypergeometric Distribution

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) =

04

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5

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6

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4

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1 2 3

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0.2

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0.6

The Hypergeometric DistributionSuppose 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.

,)(

n

N

b

B

g

G

gXP

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) = n where = G / N)

SD(X) =

Var(X) =

)1(1

nN

nN

)1(1

nN

nN

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?).

1

N

nN

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?