Winter 2007 Solutions to Quiz 2 Page: 1
Mathematical Statistics MATH 183
(Each problem is worth 25 points.)
Problem 1 This problem can be done using combinatorics, or by looking at
it as a special case of the hypergeometric distribution (m= 3, n =
4, N = 2, k = 1). I will describe how to do this problem by combi-
natorics since the next problem will also involve the hypergeometric
distribution. There are 3 vowels, and 4 consonants, so the number of
pairs with one vowel and one consonant is 3 ×4 = 12. (Notice that
order does not matter here so there is no need to multiply by 2!, e.g.
the pairs EM and M E are not counted separately) The total number
of pairs is 7
2=7×6
2= 21 since we are choosing 2 tiles out of the urn
of 7. Thus the probability is 12/21 = 4/7.
Problem 2 This problem can be modeled using the hypergeometric distribu-
tion. Since there are 21 total diamonds, we know that m+n= 21.
We want to select real diamonds, so we let n= 8, the number of real
diamonds, and let m= 13, the number of fake diamonds.
Since we want to know the expectation, it suffices to use the formula
E[X] for Xa random variable corresponding to the hypergeometric
distribution with the above parameters. Specifically,
E[X] = nN
m+n=4×8
21 = 32/21 ≈1.52
Alternatively, one can compute the expectation by calculating the sum
X
k=0..48
k13
4−k
21
4.
Problem 3 This problem can be modeled by the binomial distribution. Since
there is 500 feet of shoreline, and 30 feet of it contains the bunker, the
probability a particular missile hits is p= 30/500 = 0.06.
Thus the probability that kmissiles hit out of the volley of 25 is
25
kpk(1 −p)25−k=25
k(0.06)k(0.94)25−k.
