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Stats 241 Assignment 1
Solutions

1. A three-digit numbers is formed from the digits 0, 1, 2, 3, 4, 5 with each digit used
a) How many such digits can be formed?
Solution:        Let N = the number of three-digit numbers that can be formed from the
digits 0, 1, 2, 3, 4, 5 if each digit can only be used once = N1N2N3 where
N1 is the number of ways of selecting the first digit = 5 (0 excluded)
N2 is the number of ways of selecting the 2nd digit = 5 (0 included, first digit selected
excluded)
rd
N3 is the number of ways of selecting the 3 digit = 4 (first two digits selected excluded)

Thus N = 5.5.4 =100

b) How many of these are odd numbers?
Solution:       To count the number of outcomes in this set, assign the 3rd digit first, the
1st digit second and the second digit third. Thus
N1 is the number of ways of selecting the last digit = 3 (from 1, 3 or 5)
N2 is the number of ways of selecting the 1st digit = 4 (0 and the last digit selected
excluded)
N3 is the number of ways of selecting the 2nd digit = 4 (first and last digit selected
excluded)

Thus N = 3.4.4 =48

c) How many are greater than 330?
Solution: Let A = {x | x  300 and is a number that satisfies the conditions of part
a)}and B = {x | 300 x  329 and is a number that satisfies the conditions of part a)}.
Let C = {x | x > 330 and is a number that satisfies the conditions of part a)}. Then
n(C) = n(A) – n(B) and
n(A) = 3.5.4 = 60 (since the first digit has to be chosen from {3,4,5}) and
n(B) = 3.4 =12 (since the first two digits has to be chosen from {30,31, 32}
Thus n(C) = 60 – 12 = 48

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Stats 241 Assignment 1                                                              Solutions

2. How many distinct ways can six trees be ordered when planted in a circle?
Solution:      Since the ordering is not affected by where the first tree is planted, we
need only count the number of ways of positioning the 2nd, 3rd,4th ,5th and 6th trees
once the position of the first tree has been selected.
Thus N = N2N3N4N5N6 = 5.4.3.2.1 = 5! = 120

3. From a group of five men and three women a committee of three is selected at
random.
a) Find the probability that the committee contains two men and one woman.
Solution:
 5   3
 2   1
P[2 men and 1 woman] =     = (56) = 28
(10)(3)  15
8
 3
 
b) Find the probability that the committee contains no men.
Solution:
 5   3
 0   3
P[0 men and 3 women] =     = (56) = 56
(1)(1)     1
8
 3
 
c) Find the probability that the committee contains no women.
Solution:
 5  3
 3   0  (10)(1)   5
P[3 men and 0 women] =     = (56) = 28
8
 3
 

4. Three balls are drawn randomly without replacement from a box containing 3 red, 4
yellow, 2 white and 5 blue balls .
a) What is the probability that they are all the same colour.
Solution:
a) Let R = the event that the three balls are red.
Let Y = the event that the three balls are yellow.
Let W = the event that the threeballs are white.
Let B = the event that the three balls are red.
P[all same colour] = P[RYWB] = P[R] + P[Y] + P[W] +P[B]
3      4          5
3      3          3        (3)(2)(1)      (4)(3)(2)(1)   (10)(3)(2)(1) 15
= 14 + 14 + 0 + 14 = (14)(13)(12) + (14)(13)(12) + (14)(13)(12) = 364
                   
 3   3             3 

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Stats 241 Assignment 1                                                                 Solutions

b) What is the probability that they are all a different colour.
Solution:
Let RYW = the event that a red, yellow and white ball are selected.
Let RYB = the event that a red, yellow and blue ball are selected.
Let RWB = the event that a red, white and blue ball are selected.
Let YWB = the event that a yellow, white and blue ball are selected.

= P[RYW] + P[RYB] + P[RWB] + P[YWB]
(3)(4)(2) (3)(4)(5) (3)(2)(5) (4)(2)(5) 24+60+30+40                     154         11
= 14         +            +            +          = (14)(13)(2) = (14)(13)(2) = 26
            14          14       14 
 3          3           3        3 
5. How many integers are there between 1,000,000 and 10,000,000 in whose decimal
form no two consecutive digits are the same? (1,212,121 would be such an integer but
1,234,566 would not because the last two consecutive digits are the same.)
Solution: An integer between 1,000,000 and 10,000,000 will be a seven-digit number.
The number of ways to select the first digit will be 9 since 0 will be excluded from this
selection. The number of ways of selecting subsequent digits will also be 9, since the
digit that was excluded at the previous selection can now be included and the digit that
was selected at the previous selection must now be excluded. The number of ways this
seven-digit number can be selected is:
9999999  9 7  4782969

6. A box contains 8 red, 3 white, and 9 blue balls. Three balls are to be drawn, without
replacement. What is the probability that more blues than whites are drawn
Solution: Let Ai,j denote the event that i blue balls are selected, j white balls are selected
 9  3       8       
 i  j  3  (i  j )  .
and (3-i-j) red balls are selected. Now n(Ai,j) =                           
                    
Let A denote the event that more blue than whites are drawn.
Then A = A1,0 A2,0 A3,0 A2,1
and n(A)= n(A1,0) + n(A2,0) + n(A3,0) + n(A2,1)
 9  3  8   9  3  8   9  3  8   9  3 8 
=                   
 1  0  2   2  0  1   3  0  0   2  1  0 
               
=(9)(1)(28)+(36)(1)(8)+(84)(1)(1)+(36)(3)(1) = 732.
 20
n(S) = no. of ways of choosing 3 balls from 20 =   = 11403
 
732
Hence P[A] =
1140

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Stats 241 Assignment 1                                                                    Solutions

7. Three integers are selected at random from the set (1,2,...,10). What is the probability
that the largest of these is 5?
10 
Solution: N = the number of ways 3 integers can be selected from 10 =   = 120.
3
 
 4
The number of ways 3 integers so that the largest is 5 = (1)   =(1)(6)=6.
 2
 
6   1
Hence the probability that the largest of these is 5 is    
120 20

8. Find the probability that a poker hand contains three of a kind (exactly 3 cards one
face value and 2 cards of different face values).
Solution: N = N1N2N3N4 ,where
N1 = the number of ways the denomination of the three of a kind can be selected = 13
 4
N2 = number of ways 3 suits can be selected for the three of a kind =    4
 3
N3 = number of ways two non-matching denomination of the remaining 2 cards can be
 12  12 11
selected              66 , and
2      2 1
N4 = number of ways the suits of the remaining 2 cards can be selected= 4  4  16 ,

Hence the probability of three of a kind is
13  4  66  16 13  4  66  16    5  4  3  2.1     22  4     88
                                                           0.021128
 52              52  51  50  49.48              17  5  49 4165
5
 

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