2009-I-14

$\begin{array}{cl}
= & \dfrac{1050\times 8 + \cdots + 1450\times 10}{8 + 50 + 42 + 10 + 10} \\
= & 1220 \text{ hours}
\end{array}$

The estimated mean lifetime of Brand $B$

$\begin{array}{cl}
= & \dfrac{1050\times 4 + \cdots + 1450 \times 28}{4 + 12 + 40 + 36 + 28} \\
= & 1310 \text{ hours}
\end{array}$

Since the estimated mean lifetime of Brand $B$ is longer than that of Brand $A$, then light bulbs of Brand $B$ is likely to have a longer lifetime.

1. Note that in Brand $A$, there are $120$ light bulbs and $100$ light bulbs have lifetime less than $1300$ hours. Then the required probability

   $\begin{array}{cl}
   = & \dfrac{100}{120} \\
   = & \dfrac{5}{6}
   \end{array}$

2. The required probability

   $\begin{array}{cl}
   = & 1 – \text{P(all are acceptable)} \\
   = & 1 – \dfrac{100}{120} \times \dfrac{99}{119} \\
   = & \dfrac{73}{238}
   \end{array}$

3. Note that in Brand $B$, there are $120$ light bulbs and $56$ light bulbs have lifetime less than $1300$ hours. For method 1,

   $\begin{array}{cl}
   & \text{P(at least one good light bulb)} \\
   = & \dfrac{1}{2} \times \dfrac{73}{238} + \dfrac{1}{2} \times \left(1 – \dfrac{56}{120} \times \dfrac{55}{119}\right) \\
   = & \dfrac{779}{1428}
   \end{array}$

   For method 2,

   $\begin{array}{cl}
   & \text{P(at least one good light bulb)} \\
   = & 1 – \dfrac{100}{120}\times \dfrac{56}{120} \\
   = & \dfrac{11}{18} \\
   > & \dfrac{779}{1428}
   \end{array}$

   Therefore, method 2 has a greater chance of choosing at least one good light bulb.