Diketahui	N=10 2 lampu gagal sebelum waktu \(t = 3\)
Rumus yang digunakan	\({F_T}\left( t \right) = P\left( {T \le t} \right) = 1 – {e^{ – \lambda t}}\)
Proses Pengerjaan	\({F_T}\left( 3 \right) = P\left( {T \le 3} \right) = 1 – {e^{ – 3\lambda }}\) \(\frac{2}{{10}} = 1 – {e^{ – 3\lambda }}\) \({e^{ – 3\lambda }} = 0,8\) \(– 3\lambda = \ln (0,8)\) \(\lambda = \frac{{ – 0,223143}}{{ – 3}}\) \(= 0,07438\)
Jawaban	a. 0,0744