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Some practice for the Binomial and Hypergeometric Distributions
- Consider a hat with four slips of paper labeled A and six slips of paper labeled B. Your goal is to count the number of A’s drawn.
a) You are drawing three slips of paper one at a time from the hat and return them between draws. Construct a tree diagram and use it to determine the probabilities of each possible outcome.
P(3 A’s) = 0. P(2 A’s) = 0.096+0.096+0.096=0. P(1 A ) = 0.144+0.144+0.144=0. P(0 A’s) = 0.
b) You are drawing three slips of paper one at a time from the hat without returning them between draws. Construct a tree diagram and use it to determine the probabilities of each possible outcome.
P(3 A’s) = 1/30 ≈ 0. P(2 A’s) = 1/10+1/10+1/10 = 0. P(1 A ) = 1/6+1/6+1/6 = 0. P(0 A’s) = 1/6 ≈ 0.
c) Using the definition of a hypergeometric and binomial experiment, identify each of a and b as either hypergeometric or binomial.
A is binomial because: the 3 trials are identical (the hat is the same), each trial has two outcomes (A or B), the probability of a “success” is constant (always 6 out of 10 A’s), and the trials are independent.
B is hypergeometric because: The population is finite (10) with a fixed number of successes (4 A’s), and the sample (of size 3) is taken without replacement).
d) Match the values you got from a and b with the values you would get by using the appropriate p.m.f.
For a,
[ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
0 30 3
1 31 2
2 32 2
3 33 3
⎟⎟ =^ =
−
−
−
−
P X
PX
PX
PX
For b,
[ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
P X
P X
P X
P X
- A class of 24 students contains 15 women and 9 men. Six of these students are to be chosen at random for a demonstration in front of the class.
a) How many ways can 1 man be chosen from a group of 9? (9 choose 1)=9! / 1!8!=
b) How many ways can 5 women be chosen from a group of 15? (15 choose 5)=15! / 10!5! = 15·14·13·12·11/5·4·3·2·1=
c) How many ways can a group of 1 man and 5 women be chosen from a larger group of 9 men and 15 women? We need to use the multiplication principal here, so (9 choose 1)(15 choose 5)=(9)(3003)=27,
d) How many ways can six students be chosen from a group of 24? (24 choose 6)= 24·23·22·21·20·19/6·5·4·3·2·1=134,
e) What is the probability that the group of six students chosen will have only 1 man? (Does this match the probability from the hypergeometric p.m.f.?) Each of the random selections of 6 students out of 24 has the same probability, so it is 27,027/134,597=0.2008. Looking at the values in c and d, this is a binomial with population size 24, 9 “defectives”, and sample size six.
- Consider a population of size N containing Np defectives. A sample of size n is to be chosen without replacement.
a) Is this a hypergeometric or binomial? Hypergeometric.
b) Give the mean and standard deviation of the number of defectives contained in the sample.
We are using N for the population size (usually n), n for the sample size (usually m), and Np for the number of defectives (usuall r). So:
c) What happens to these values as N increases? They become identical to the values from the binomial distribution.
n
r
μ X m ⇒ np
n 1
n m n
n r n
r
σ X m
N
N n np p
