Epple公司生产大量的E-phone。已知生产的E-phone中有 1% 是次品。

1. Define the random variable and its distribution

Let $X$ be the number of defective E-phones in a sample of 10.
$X \sim \mathrm{B}(10, 0.01)$

2. Formulate and calculate the required probability

The probability that at least 1 but less than 3 are defective is $\mathrm{P}(1 \le X < 3)$.
$\mathrm{P}(1 \le X \le 2) = \mathrm{P}(X = 1) + \mathrm{P}(X = 2)$
$= \binom{10}{1}(0.01)^1(0.99)^9 + \binom{10}{2}(0.01)^2(0.99)^8$
$= 0.091352 + 0.004152$
$= 0.095504 \approx 0.0955 \text{ (to 3 s.f.)}$

1. Translate the condition into an equivalent combination of events

For the 10th E-phone selected to be the 8th not defective E-phone, the first 9 E-phones selected must contain exactly $7$ not defective E-phones (and therefore $2$ defective E-phones), and the 10th E-phone must be not defective.

2. Calculate the probability using the binomial distribution and multiplication rule

$\begin{aligned} P(\text{required event}) &= P(\text{exactly } 7 \text{ not defective in first } 9) \times P(\text{10th is not defective}) \\ &= \binom{9}{7}(0.99)^7(0.01)^2 \times (0.99) \end{aligned}$

3. Evaluate the probability to 3 significant figures

$\begin{aligned} P(\text{required event}) &= 36 \times (0.99)^8 \times (0.01)^2 \\ &= 0.00332188\dots \\ &\approx 0.00332 \text{ (to 3 s.f.)} \end{aligned}$

1. Define the random variable and its distribution.

Let $Y$ be the number of defective E-phones in a carton of 24.
Since each E-phone has a $1\%$ probability of being defective, $Y$ follows a binomial distribution:
$Y \sim \mathrm{B}(24, 0.01)$

2. Calculate the probability of the carton being rejected using the complement rule.

The carton is rejected if there are at least two defective phones, so we need $\mathrm{P}(Y \ge 2)$.
Using the complement rule:
$\mathrm{P}(Y \ge 2) = 1 - \mathrm{P}(Y \le 1)$
$= 1 - [\mathrm{P}(Y = 0) + \mathrm{P}(Y = 1)]$
$= 1 - \left[ \binom{24}{0}(0.01)^0(0.99)^{24} + \binom{24}{1}(0.01)^1(0.99)^{23} \right]$
$= 1 - (0.785678 + 0.190467)$
$= 0.023855 \dots$
$\approx 0.0239 \text{ (to 3 s.f.)}$

1. Define the distributions for the total number of defective phones and approximate them using the normal distribution.

Let $X$ be the total number of defective E-phones.
Total number of E-phones $= 250 imes 24 = 6000$
$X \sim \text{B}(6000, 0.01)$
Since $n$ is large, $np = 60 > 5$ and $n(1-p) = 5940 > 5$, $X$ can be approximated by a normal distribution:
$X \sim \text{N}(60, 59.4)$

Let $Y$ be the total number of defective Galaxy phones.
Total number of Galaxy phones $= 500 imes 20 = 10000$
$Y \sim \text{B}(10000, 0.015)$
Since $n$ is large, $np = 150 > 5$ and $n(1-p) = 9850 > 5$, $Y$ can be approximated by a normal distribution:
$Y \sim \text{N}(150, 147.75)$

2. Formulate the distribution for the difference between the two variables.

Let $W = Y - X$.
Assuming the number of defective E-phones and Galaxy phones are independent:
$\text{E}(W) = \text{E}(Y) - \text{E}(X) = 150 - 60 = 90$
$\text{Var}(W) = \text{Var}(Y) + \text{Var}(X) = 147.75 + 59.4 = 207.15$
Therefore, $W \sim \text{N}(90, 207.15)$.

3. Apply the continuity correction and calculate the required probability.

We need to find the probability that $Y - X \ge 100$.
Applying the continuity correction for the discrete variables:
$\text{P}(Y - X \ge 100) \approx \text{P}(W \ge 99.5)$
$= \text{P}\left(Z \ge \frac{99.5 - 90}{\sqrt{207.15}}\right)$
$= \text{P}(Z \ge 0.660056...)$
$= 0.25461... \approx 0.255 \text{ (to 3 s.f.)}$