Pembahasan Soal Ujian Profesi Aktuaris
Institusi
:
Persatuan Aktuaris Indonesia (PAI)
Mata Ujian
:
Probabilitas dan Statistika
Periode Ujian
:
Juni 2016
Nomor Soal
:
28
SOAL
Misalkan \({X_1},{X_2},{X_3}\) ialah peubah acak yang identik dan saling bebas yang mana memiliki fungsi kepadatan peluang \(f(x) = \exp ( – x)\,,\,0 < x < \infty \,,\,dan\,f(x) = 0,\,untukx\) lainnya. Hitung \(P({X_1} < {X_2}|{X_1} < 2{X_2})\)
1/6
4/7
2/3
1/5
3/8
Diketahui
\(f(x) = \exp ( – x)\)
Step 1
\(f({x_1},{x_2}) = f({x_1})f({x_2})\)
\(f({x_1},{x_2}) = {e^{ – {x_1}}}{e^{ – {x_2}}}\,,\,0 < {x_1},{x_2} < \infty \)
Step 2
Step 3
\(P({X_1} < {X_2}) = \int\limits_0^\infty {\int\limits_0^{{x_2}} {{e^{ – {x_1}}}{e^{ – {x_2}}}} } \,d{x_1}d{x_2}\)
\(P({X_1} < {X_2}) = \int\limits_0^\infty {{e^{ – {x_2}}}\left( { – 1} \right)\left( {{e^{ – {x_2}}} – 1} \right)} d{x_2}\)
\(P({X_1} < {X_2}) = \int\limits_0^\infty {\left( {{e^{ – {x_2}}} – {e^{ – 2{x_2}}}} \right)} d{x_2}\)
\(P({X_1} < {X_2}) = \left( { – 1(0 – 1)} \right) – \left( { – \frac{1}{2}(0 – 1)} \right)\)
\(P({X_1} < {X_2}) = 1 – \frac{1}{2}\)
\(P({X_1} < {X_2}) = \frac{1}{2}\)
\(P({X_1} < 2{X_2}) = \int\limits_0^\infty {\int\limits_0^{2{x_2}} {{e^{ – {x_1}}}{e^{ – {x_2}}}} } \,d{x_1}d{x_2}\)
\(P({X_1} < 2{X_2}) = \int\limits_0^\infty {{e^{ – {x_2}}}\left( { – 1} \right)\left( {{e^{ – 2{x_2}}} – 1} \right)} d{x_2}\)
\(P({X_1} < 2{X_2}) = \int\limits_0^\infty {\left( {{e^{ – {x_2}}} – {e^{ – 3{x_2}}}} \right)} d{x_2}\)
\(P({X_1} < 2{X_2}) = \left( { – 1(0 – 1)} \right) – \left( { – \frac{1}{3}(0 – 1)} \right)\)
\(P({X_1} < 2{X_2}) = 1 – \frac{1}{3}\)
\(P({X_1} < 2{X_2}) = \frac{2}{3}\)
\(P({X_1} < {X_2}|{X_1} < 2{X_2}) = \frac{{\left( {\frac{1}{2}} \right)}}{{\left( {\frac{2}{3}} \right)}}\)
\(P({X_1} < {X_2}|{X_1} < 2{X_2}) = \frac{3}{4}\)
Jawaban Tidak tersedia pada pilihan
\(P({X_1} < {X_2}|{X_1} < 2{X_2}) = \frac{{P({X_1} < {X_2})}}{{P({X_1} < 2{X_2})}}\)