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第1章
場合の数と確率
第1節場合の数
1
集合の要素の個数（第3回）
復習
問題集
𝑛(U)=100
𝑛(A ∩ B)=15
205
n(A∪B)=70
U (100)
A
(55)
(33) B
𝑛(A ∩ B)=40
(40)
(1) 𝑛(A ∩ B) =n(A∪ B) =100-70=30
(2) 𝑛(A ∩ B) =70-(40+15) =15
(3) 𝑛(A) =40+15 =55
(4) 𝑛(B) =15+18 =33
(15)
(15)
(30)
例
U (100)
1から100までの整数のうち、
A (33)
A =｛ n | n は3の倍数｝ B =｛ n | n は4の倍数｝
C =｛ n | n は5の倍数｝
𝑛(A) =100÷3 =33
𝑛(C) =100÷5 =20
𝑛(B) =100÷4 =25
(25)
B
3,4,5のうちすくなくとも1つで割り切れる数の個数を求めよ。
𝑛(A∪B∪C)
(20)
C
U (100)
例
A (33)
A ∩B =｛ n | n は12の倍数｝
𝑛(A ∩ B) =100÷12 =8
(1)
B ∩C =｛ n | n は20の倍数｝
𝑛(B ∩ C) =100÷20 =5
C ∩A =｛ n | n は15の倍数｝
𝑛(C ∩ A) =100÷15 =6
A ∩ B ∩C
=｛ n | n は60の倍数｝
𝑛(A ∩ B ∩ C) =100÷60 =1
(6)
(8)
(25)
B
(20)
(5)
C
n(A∪B∪C)= n(A) +n(B) +n(C) -n(A∩ C) -n(B∩ C) -n(C∩ A) +n(𝐀 ∩ 𝐁 ∩C)
U
A
A
A
U
B
B
U
U
A
B
U
U
A
C
B
C
A
C
B
C
n(A∪B∪C)
= n(A) +n(B) +n(C)
-n(A∩ C) -n(B∩ C) -n(C∩ A)
+n(A ∩ B ∩C)
U
例
A (33)
A∪B∪C =｛ n | n 3,4,5のうちすくなくとも
n(A∪B∪C)=
(6)
(8)
1つで割り切れる数｝
(1)
33 +25 +20
(25)
-8 -5 -6 +1
=60
B
(20)
(5)
C
問題演習
教科書
P10 練習2
問題集
209
（210）
（211）
解答
教科書
練習2
𝑛(A) =100÷2 =50
U (100)
𝑛(B) =100÷3 =33
A (50)
𝑛(C) =100÷5 =20
𝑛(A ∩ B) =100÷6
=16
𝑛(B ∩ C) =100÷15 =6
𝑛(C ∩ A) =100÷10 =10
𝑛(A ∩ B ∩ C) =100÷30 =3
n(A∪B∪C)
=50+33+20-16-6-10+3 =74
(10)
(16)
(3)
(33)
B
(20)
(6)
C
解答
問題集
209
U
A =｛ 1, 2, 3, 4, 6, 8, 12, 24, 48 }
𝑛(A) = 10
B =｛ 1, 3, 5, ・・・, 29 }
𝑛(B) = 15
C =｛ 1, 2, 3, 6, 9, 18, 27, 54 }
(1) A∩ B =｛ 1, 3 }
A (10)
𝑛(C) = 8
𝑛(A ∩ B) = 2
B∩ C =｛ 1, 3, 9, 27 }
𝑛(B ∩ C) = 4
C∩ A =｛ 1, 2, 3, 6 }
𝑛(C ∩ A) = 4
(4)
(2)
(15)
B
(8)
(4)
C
解答
問題集
209
U
A =｛ 1, 2, 3, 4, 6, 8, 12, 24, 48 }
𝑛(A) = 10
B =｛ 1, 3, 5, ・・・, 29 }
𝑛(B) = 15
C =｛ 1, 2, 3, 6, 9, 18, 27, 54 }
(2) A∩ B ∩ C =｛ 1, 3 }
=10+15+18-2-4-4+2 =25
(4)
(2)
𝑛(C) = 8
𝑛(A ∩ B ∩ C) = 2
(3) 𝑛(A ∩ B ∩ C)
A (10)
(2)
(15)
B
(8)
(4)
C
解答
問題集
210
𝑛(U)=200
𝑛(A) =200÷3 =66
U (200)
𝑛(B) =200÷5 =40
A (66)
𝑛(C) =200÷8 =25
(1)
𝑛(B ∩ C) =200÷40 =5
(40)
𝑛(C ∩ A) =200÷24 =8
B
𝑛(A ∩ B ∩ C) =200÷120 =1
n(A∪B∪C)
=66+40+25-13-5-8+1
(8)
(13)
𝑛(A ∩ B) =200÷15 =13
=106
(25)
(5)
C
解答
問題集
210
U (200)
A (66)
(2) n(A∪B∪C) =200-106 =94
(3) n((A ∩ B) ∩ C)
= n(A∪B∪C)-𝑛(C)
=106-25
=81
(8)
(13)
(1)
(40)
B
(25)
(5)
C
解答
問題集
211
U
(1) n(A∪C)=78
A
A (65)
(65)
(x)
C
(14)
(11)
(11)
(40)
n(A∪C) = n(A) + n(C) - 𝑛(A ∩ C)
55 = 65 + x - 11
∴ x = 24
B
(24)
C
解答
問題集
211
U
(2) n(A∪C)=78
B
A (65)
(40)
(24)
C
(11)
(14)
(y)
(40)
n(B∪C) = n(B) + n(C) - 𝑛(B ∩ C)
55 = 40 + 24 - y
∴ y=9
B
(24)
(9)
C
解答
問題集
211
U
(2) n(A∪B∪C)
A (65)
= n(A) +n(B) +n(C)
(11)
(14)
-n(A∩ C) -n(B∩ C) -n(C∩ A)
(4)
+n(A ∩ B ∩C)
(40)
99 = 65+40+24-14-9-11+n(A ∩ B ∩C)
n(A ∩ B ∩C) = 99-129+34 =4
B
(24)
(9)
C
解答
問題集
211
U
(3) n(A ∩ (B∪C)) = 65-(14+11-4) = 44
A (65)
n(B ∩ (C∪A)) = 40-(14+9-4) = 21
(44)
n(A ∩ (B∪C)) = 65-(14+11-4) = 8
∴ 44+21+8=73
(11)
(14)
(4)
(40) (21)
B
(8)
(9)
(24)
C

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