A3. (a) A police department in a small city consists of 10 officers. If the department
policy is to have 5 of the officers patrolling the streets, 2 of the officers working
full time at the station, and 3 of the officers on reserve at the station, how many
different divisions of the 10 officers into the 3 groups are possible?
(b) It is known that screws produced by a certain company will be defective with
probability 0.01 , independently of each other. The company sells the screws in
packages of 10 and offers money back guarantee that at most 1 of the 10 screws
is defective. What proportion of packages sold must the company replace?

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**(a) Division of Officers**

We need to divide 10 officers into three groups of sizes 5, 2, and 3. First, choose 5 officers for the patrol group, then 2 of the remaining for full-time station duty, and the remaining 3 will be on reserve.

1. Choose 5 officers from 10:
   $$
   \binom{10}{5}
   $$
2. Choose 2 officers from the remaining 5:
   $$
   \binom{5}{2}
   $$
3. The remaining 3 officers automatically form the reserve group:
   $$
   \binom{3}{3}=1
   $$

Thus, the total number of different divisions is:
$$
\binom{10}{5} \times \binom{5}{2} \times \binom{3}{3} = \binom{10}{5} \times \binom{5}{2}.
$$

Calculating,
$$
\binom{10}{5}=252 \quad \text{and} \quad \binom{5}{2}=10.
$$

Thus,
$$
252 \times 10 = 2520.
$$

**(b) Replacement of Screw Packages**

Each screw produced is defective independently with probability 0.01. A package contains 10 screws. The company offers a money-back guarantee if at most 1 screw in the package is defective. Therefore, the company must replace a package if there are 2 or more defective screws.

Let $$ X $$ be the number of defective screws in a package. Then,
$$
X \sim \text{Binomial}(10, 0.01).
$$

The probability that a package meets the guarantee (i.e., at most one defective screw) is:
$$
P(X \leq 1) = P(X=0) + P(X=1).
$$

Calculate $$ P(X=0) $$:
$$
P(X=0) = \binom{10}{0}(0.01)^0(0.99)^{10} = (0.99)^{10}.
$$

Calculate $$ P(X=1) $$:
$$
P(X=1) = \binom{10}{1}(0.01)^1(0.99)^9 = 10 \times 0.01 \times (0.99)^9.
$$

Thus, the total probability that a package is acceptable is:
$$
(0.99)^{10} + 10 \cdot 0.01 \cdot (0.99)^9.
$$

The proportion of packages that must be replaced (i.e., having 2 or more defective screws) is:
$$
1 - \left[(0.99)^{10} + 10 \cdot 0.01 \cdot (0.99)^9\right].
$$

This is the exact answer. For an approximate numerical value, note that:
$$
(0.99)^{10} \approx 0.9044 \quad \text{and} \quad 10 \cdot 0.01 \cdot (0.99)^9 \approx 0.0914,
$$
so,
$$
(0.99)^{10} + 10 \cdot 0.01 \cdot (0.99)^9 \approx 0.9044 + 0.0914 = 0.9958.
$$

Hence, the proportion of packages that must be replaced is approximately:
$$
1 - 0.9958 = 0.0042.
$$

That is, about $$0.42\%$$ of the packages sold must be replaced.
**(a) Division of Officers** There are 2,520 different ways to divide the 10 officers into groups of 5 patrolling, 2 at the station, and 3 on reserve. **(b) Replacement of Screw Packages** About 0.42% of the packages sold must be replaced due to having 2 or more defective screws.

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