(超急~唔該!)數一題~~~~~20分20分

1)在3,5,6,8中隨意抽出兩個號碼(抽出不放回)

b)求兩數之和為奇數的概率
Given 奇數 (3, 5) = 2; 偶數 (6, 8) = 2
If first 抽出號碼是奇數 (3, 5) , Pb1 = 2 / 4 =
In order that 兩數之和為奇數, second 抽出號碼 must be偶數 (6, 8) , Pb2 = 2 / (4 – 1) = 2 /3
Therfore, 兩數之和為奇數的概率 Pb3 = Pb1 * Pb2 = * 2/3 = 1/3
Similarly, If first 抽出號碼是偶數 (6, 8) , Pb4 = 2 / 4 =
In order that 兩數之和為奇數, second 抽出號碼 must be奇數 (3, 5) , Pb5 = 2 / (4 – 1) = 2 /3
Therfore, 兩數之和為奇數的概率 Pb6 = Pb4 * Pb5 = * 2/3 = 1/3
So, 兩數之和為奇數的概率 = Pb3 + Pb6 = 1/3 + 1/3 = 2 /3 (ans.)

c)求兩數之積為偶數的概率
Given 奇數 (3, 5) = 2; 偶數 (6, 8) = 2
If first 抽出號碼是奇數 (3, 5) , as奇數 * 奇數 = 奇數, Pc1 = 0 / 4 = 0
If first 抽出號碼是偶數 (6, 8) , Pc2 = 2 / 4 =
In order that 兩數之積為偶數, second 抽出號碼 must also 為 偶數 (remaining one of 6, 8) , Pc3 = 1 / (4 – 1) = 1 /3
Therfore, 兩數之積為偶數的概率 Pc4 = Pc2 * Pc3 = * 1/3 = 1/6
So, 兩數之積為偶數的概率 = Pc1 + Pc4 = 0 + 1/6 = 1 /6 (ans.)
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