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Unformatted text preview: 0×10 = 100 .
10
1
1
P(letters are identical and digits are equal) = 100 × 1 = 400 .
4
P(the letters are formed by B and D) = P({B,D} or {D,B}) =
2
1
4×4 = 8 .
5
P(the 3digit number is even) = 10 = 1 .
2
P(the letters are formed by B and D, and the 3digit number is even)
1
1
1
= 8 × 2 = 16 . Problem 2 Problem 2. Three boys and four girls line up randomly in a row.
What is the probability that
1
2
3
4 all boys are on one side?
boys and girls are in alternate positions?
no 2 boys sit next to each other?
the boy named John sits on the right of the girl named Mary (assuming
all boys and girls have diﬀerent names)? Solution to Problem 2 Solution.
1
2
3 4 2
7
=.
3
35
1
P(boys and girls are in alternate positions) = 35 .
The number of ways of arranging in this way is 5 = 10 ( Set the girls
3
as GGGG, put the three boys in to the interspace of the girls). Thus
P(no two boys sit next to each other) = 10 = 2
35
7
6 × 5! 1
P(John sits on the right of Mary)=
=.
7!
7
P(all boys on one side)= 2/ Problem 3
Problem 3. There are 4 red balls, 3 yellow balls, 2 blue balls and 1
white balls. In how many distinct ways can the balls be arranged in a
circle?
Solution.
First we consider the case of arranging the balls in a row. It is a
problem of arrangement of n objects (10 balls), with r
distinguishable types (4 colors).
10
10!
=
4! × 3! × 2! × 1!
4, 3, 2, 1 = 12600 We then understand that for each arrangement of balls in a circle, we
can derive 10 distinct arrangements in a row, each by choosing a
diﬀerent ball to be the ﬁrst in a row. The number of ways the balls
can be arranged in a row is 10 times of that they can be arranged in a
circle. Thus, the answer is:
12600
10 = 1260 Problem 4 Problem 4. (Exam Fall 2008) Three candidates, A, B and C,
participate in an election in which eight voters will cast their votes.
The candidate who receives the absolute majority, that is at least ﬁve,
of the votes will win the election. All voters must vote for one and
only one candidate and cannot abstain.
Suppose the voting is done by a...
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 Fall '08
 SMSLee
 Statistics, Probability

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