設 paper record 正確的機率為P(p), 則不正確的機率為 P(p[sup]c[/sup]) = 1 - P(p)
設 electric record 正確的機率為P(e), 則不正確的機率為 P(e[sup]c[/sup]) = 1 - P(e)
依據題意:
60 percent of the policies having incorrect paper record have incorrect electric record
P(e[sup]c[/sup]|p[sup]c[/sup]) = 60% = 0.6
==> P(e[sup]c[/sup] 交集 p[sup]c[/sup]) / P(p[sup]c[/sup]) = 0.6
==> 0.6 * P(p[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup])
.........<I>
75 percent of the policies having incorrect electric record have incorrect paper record
P(p[sup]c[/sup]|e[sup]c[/sup]) = 75% = 0.75
==> P(p[sup]c[/sup] 交集 e[sup]c[/sup]) / P(e[sup]c[/sup]) = 0.75
==> 0.75 * P(e[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup])
.........<II>
3 percent of all the policies have both incorrect paper and incorrect electric records
P({p 交集 e}[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 3% = 0.03
.........<III>
並且 P({p 交集 e}[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 1 - P({p 聯集 e}) = 0.03
==> P({p 聯集 e}) = 0.97
由<I><III>可得知
0.6 * P(p[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 0.03
==> 0.6[1 - P(p)] = 0.03
==> P(p) = 0.95
由<II><III>可得知
0.75 * P(e[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 0.03
==> 0.6[1 - P(e)] = 0.03
==> P(e) = 0.96
題目問的是
If we randomly pick out one policy, what is the probability that it is one having both correct paper and correct electric record
意即求 P(p 交集 e) = ?
由貝式定理得知
P({p 聯集 e}) = P(p) + P(e) - P(p 交集 e)
0.97 = 0.95 + 0.96 - P(p 交集 e)
P(p 交集 e) = 0.94
這題題目有點繞...把文式圖畫出來應該會比較好理解...
