STAT 230
Solutions to Problems: Chapter 3
Note:
Questions who solutions are not here are answered sufficiently well in the text.
3.1
a)
Notice that you can select the six digits in 7
(6)
ways. Now such a number is even if it
ends with 2, 4, 6, or 8. Let’s turn the problem around and imagine we build the
number from right to left.
To obtain an even number then we must start with 2, 4, 6,
or 8 . That is we can pick the first digit 4 ways, then the next in 6 then the next in 5
and so on. This results in 4*6*5*4*3*2 ways to pick an even number. The probability
then is
7
4
2
3
4
5
6
7
2
3
4
5
6
4
=
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
.
Another way:
Again, think of building the number from right to left. The probability
that you’ll select an even digit first is 4/7 , since 4 of the 7 digits available are even.
b) You can have the “2” appear in 5 positions. The “3” is then determined with no
choice and you can fill in the remaining 4 positions as usual with the remaining 5
digits.
. The number of ways of building such a number is then 5*5*4*3*2 so the
probability is
42
5
2
3
4
5
6
7
2
3
4
5
5
=
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
c)
Considerably harder, must be done exhaustively. Notice first though that the other 4
spots can still be filled in 5·4·3·2 ways. We need to find how many ways the 2 and 3
can be placed, then multiply by that.
If the 2 is the first digit, the 3 has 4 possible locations (remember they can’t be
consecutive, so “3” cannot appear 1
st
or 2
nd
). If 2 is the second digit, 3 has 3 locations
available. It’s not hard to see then that the total number of locations for the 2 and 3
together are: 4+3+2+1 = 10. The probability then is
21
5
2
3
4
5
6
7
2
3
4
5
10
=
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
3.2 –
3.3 There are 6
4
ways in total they can get off.
a)
Two stops will have noone get off, select those stops in
15
2
6
=
ways. Then there
are 4 stops the first passenger can get off at, 3 for the second … for a total of 4·3·2 =
24 ways. Probability is
18
5
6
)
24
(
15
4
=
.
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View Full Documentb) Select the two stops
15
2
6
=
ways. Then select two people to get off at the first stop
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 Spring '10
 Mr.Lushman
 Arithmetic progression, #, $3000

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