Homework 1 Solutions - Statistics Department
Homework 1 Solutions
1. Ross 1.1 p. 16
a) Since repeats are allowed in this scenario, there are 26 possibilities are each of the first 2 places and 10 possibilities for each of the remaining 5 places; hence, there are [pic] different 7-place license plates.
b) Now, repeats are not allowed, so while the first letter can be any of 26 possibilities, the second letter can only be chosen from the remaining 25; the same applies for the 5 places that hold numbers. Hence, we have [pic] different 7-place license plates.
2. Ross 1.10 p. 16
a) With 0 restrictions on the seating arrangement, there are [pic] different ways to seat 8 people in a row.
b) If we require that persons A and B sit next to each other, then we have 14 different seat pairings for A and B satisfying this restriction: (i.e., (A1, B2), (A2, B3), …, (A7, B8) and their flipped counterparts (B1, A2), (B2, A3), …, (B7, A8)). The remaining 6 individuals for each of these pairings can be arranged in [pic] different ways. Thus, if A and B must sit next to each other, then we have [pic] possible seating arrangements.
c) We now have 4 men and 4 women, and we cannot have 2 members of the same gender sitting next to each other. So, for the first seat, we have 8 possible choices. The second seat has 4 possible choices (the 4 corresponding to the opposite gender from the individual chosen in the first seat); the third and fourth seats both have 3 possible choices; the fifth and sixth seats both have 2 possible choices; and one choice exists for both the seventh and eighth seat. Hence, we have [pic] different ways to alternate the males and females.
d) With 5 males needing to be seated consecutively, we have 4 different seat sequences that satisfy this ((1, …, 5), (2, …, 6), (3, …, 7), (4, …, 8)). For each of these 4 sequences, there are [pic] different ways to seat the 5 males, and [pic] different ways to seat the 3 females. So, we have [pic] different ways to seat 5 males next to each other.
e) There are 4 married couples and the couples must sit together. Instead of looking at each seat as a unit, we instead consider each pair of seats a unit (so seats 1 and 2 form slot 1, seats 3 and 4 form slot 2, and so on). This gives us 4 slots in which we can arrange the 4 couples. So, there are [pic] different ways to arrange 4 couples. This ignores that each slot can have 2 possible configurations of husband and wife; so, for each of the 24 arrangements, we have [pic] possible sequences of individuals. Thus, overall, there are [pic] different ways to seat 4 different couples in a row while keeping each couple seated together.
3. Ross 1.19 p. 17
a) We have 8 women and 6 men, and we wish to form a committee of 3 men and 3 women. If 2 of the men refuse to serve together, then we have
Committees with neither man: [pic]
Committees with one of the men: [pic]
Therefore, there are a total of [pic] committees possible when 2 of the men refuse to serve together.
b) Now, 2 of the women refuse to serve together. Treating it the same as in (a), we have
Committees with neither woman: [pic]
Committees with one of the two: [pic]
Therefore, there are a total of [pic] committees possible when 2 of the women refuse to serve together.
c) Now, 1 of the women and 1 of the men refuse to serve together. This gives us
Committees with neither: [pic]
Committees with M but not F: [pic]
Committees with F but not M: [pic]
This gives us a total of [pic] possible committees where 1 of the men and 1 of the women refuse to serve together.
4.
a) Ross 1.21 p. 17
We have a total of 7 steps, of which 4 are to the right and 3 are up. Hence, we have [pic] possible paths from A to B.
b) Ross 1.22 p. 17
To go from A to the circled point requires 4 steps: 2 right, 2 up. So, there are [pic] different paths by which one can go from A to the circled point. To go from the circled point to B requires 3 steps: 2 right, 1 up. So, there are [pic] possible paths from the circled point to B. Thus, there are a total of [pic] different paths from A to B that go through the circled point.
5. Ross 2.1 p. 55
a) With replacement, the sample space is {(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)}.
b) Without replacement, the sample space is {(R,B), (R,G), (B,R), (B,G), (G,R), (G,B)}.
6. Ross 2.6 p. 56
a) The sample space for this experiment is {(1,g), (1,f), (1,s), (0,g), (0,f), (0,s)}.
b) If A is the event that the patient is in serious condition, the outcomes in A are {(1,s), (0,s)}.
c) If B is the event that the patient is uninsured, then the outcomes in B are {(0,g), (0,f), (0,s)}.
d) The event [pic] is the event that the patient is either insured or in serious condition, so the outcomes in this event are {(1,g), (1,f), (1,s), (0,s)}.
7.
There are a total of [pic] possible outcomes for the months 6 people were born in. Within these [pic] possible outcomes, there are [pic] for which no two people are born in the same month. Hence, the probability that at least 2 people have a birthday in the same month is just [pic].
One possible R code for simulating this is as follows (two runs of this produced proportions of 0.7724 and 0.7799, so it’s in the right neighborhood):
# R program for birthday problem
# Simulate the probability that at least two of k people
# have a birthday in the same month
k ................
................
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