Quiz 7
Name:
Exam, Form: A
Section 1. Short Answer:
Throughout this quiz, you may leave your answer in terms of binomial coefficients, exponents,
factorials, powers (of e), and fractions.
1. A coin is rigged to show heads with probability 1/5. It is tossed over and over.
If we use the Poisson distribution to estimate the number of heads in 30 tosses,
what should be used for λ?
We can estimate the binomial probability b(30, 1/5, k) with a Poisson distribution whose parameter t
is 1 (as in 1 duration of 30 tosses) and
λ = n · p = 30 · (1/5) = 6 .
2. What does the Poisson distribution give as its estimate for the probability of getting
exactly 4 heads in 30 tosses?
64 −6
λ4 −λ
e =
e = 54e−6 ≈ 0.1339.
4!
24
3. What does the Poisson distribution give as its estimate for the probability of getting
at least 2 heads in 30 tosses?
The complement is “zero or one heads,” so the probability of at least 2 heads in 30 tosses is 1 minus
the probability of zero heads minus the probability of one head:
1−
60 −6 61 −6
e − e = 1 − e−6 − 6e−6 = 1 − 7e−6 ≈ 0.9826.
0!
1!
4. An urn contains 4 red marbles and 5 white marbles (9 marbles total). An experiment consists of
drawing 3 marbles without replacement, so we draw a subset of 3 marbles. Use the appropriate
discrete distribution to find the probability that there are 2 red marbles (and 1 white marble) among
the 3 marbles that were drawn.
We can use the hypergeometric distribution. The numerator counts how many subsets of 3 marbles
have 2 red ones. To count these subsets, we split the task into two parts: choosing exactly 2 red
marbles to go into our subset, and choosing enough white marbles to fill out the rest of our subset:
(1) There are 4 red marbles, and we choose two of them to go into our subset:
4
= 6.
2
(2) We need a subset of 3 marbles, so there is one marble left to choose. It cannot be a red marble
since we want exactly two red marbles, so the remaining marble must be a white one. Thus, we
choose 1 remaining marble to fill out our subset from a choice of 5 white marbles:
5
= 5.
1
Multiply these, and then divide by the total number of ways to choose 3 marbles from 9 total (our
sample space is subsets of 3 marbles chosen from 9 total):
4
5
5
2
1
30
=
.
=
84
14
9
3