Pembahasan-Soal-Ujian-Profesi-Aktuaris-1024x481 (3) pai

Pembahasan Ujian PAI: A20 – No. 17 – Juni 2014

Pembahasan Soal Ujian Profesi Aktuaris

Institusi:Persatuan Aktuaris Indonesia (PAI)
Mata Ujian:Probabilita dan Statistika
Periode Ujian:Juni 2014
Nomor Soal:17

SOAL

Diketahui kejadian A, B dan C memenuhi hubungan sebagai berikut :

\(A \cap B'{\rm{ }} = \phi ,B \cap C'{\rm{ }} = \phi ,{\rm{ }}Pr(A’ \cap B){\rm{ }} = x,{\rm{ }}Pr(B’ \cap C){\rm{ }} = y,{\rm{ }}Pr\left( C \right){\rm{ }} = z\)

Maka Pr(A) sama dengan …

  1. x + y + z
  2. x + y – z
  3. y + z – x
  4. x – y + z
  5. z – x – y
[showhide type more_text=”Kunci Jawaban & Pembahasan” less_text=”Sembunyikan Kunci Jawaban & Pembahasan”]
Diketahui\(A \cap B'{\rm{ }} = \phi ,B \cap C'{\rm{ }} = \phi ,{\rm{ }}Pr(A’ \cap B){\rm{ }} = x,{\rm{ }}Pr(B’ \cap C){\rm{ }} = y,{\rm{ }}Pr\left( C \right){\rm{ }} = z\)
Rumus yang digunakan\(\Pr (B \cup C’) = \Pr (B) + \Pr (C’) – \Pr (B \cap C’)\) \(\Pr (A \cup B’) = \Pr (A) + \Pr (B’) – \Pr (A \cap B’)\)
Proses pengerjaan\(\Pr (B \cup C’) = \Pr (B) + \Pr (C’) – \Pr (B \cap C’)\) \(\Leftrightarrow 1 – \Pr (B’ \cap C) = \Pr (B) + \Pr (C’) – 0\) \(\Leftrightarrow 1 – y = \Pr (B) + (1 – z)\) \(\Leftrightarrow z – y = \Pr (B)\) \(\Pr (A \cup B’) = \Pr (A) + \Pr (B’) – \Pr (A \cap B’)\) \(\Leftrightarrow 1 – \Pr (A’ \cap B) = \Pr (A) + \Pr (B’) – 0\) \(\Leftrightarrow 1 – x = \Pr (A) + (1 – z + y)\) \(\Leftrightarrow z – x – y = \Pr (A)\)
Jawabane. z – x – y
[/showhide]

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