179. A factory sent 500 items to a base. The probability of an item being damaged in transit is 0.002. Find the probabilities that during transit the number of damaged items will be: a) exactly three; b) less than three; c) more than three; d) at least one.
Solution. The number $n=500$ is large, the probability $p=0.002$ is small, and the events under consideration (damage to items) are independent, so the Poisson formula applies:

$$
P_{n}(k)=\lambda k \cdot e^{-\lambda} / k!
$$

a) Let's find $\lambda$:

$$
\lambda=n p=500 \cdot 0.002=1
$$

Let's find the probability that exactly $3(k=3)$ items will be damaged:

$$
P_{500}(3)=e^{-1} / 3!=0.36788 / 6=0.0613
$$

b) Let's find the probability that fewer than three items will be damaged:

$$
\begin{gathered}
P_{500}(0)+P_{500}(1)+P_{500}(2)=\mathrm{e}^{-1}+\mathrm{e}^{-1}+e^{-1} / 2= \\
=(5 / 2) e^{-1}=(5 / 2) \cdot 0.36788=0.9197 .
\end{gathered}
$$

c) Let's find the probability $P$ that more than three items will be damaged. The events "more than three items are damaged" and "not more than three items are damaged" (let's denote the probability of this event as $Q$) are complementary, so $P+Q=1$. Therefore,

$$
P=1-Q=1-\left[P_{500}(0)+P_{500}(1)+P_{500}(2)+P_{500}(3)\right] .
$$

Using the results obtained above, we have

$$
P=1-[0.9197+0.0613]=0.019
$$

d) Let's find the probability $P_{i}$ that at least one item will be damaged. The events "at least one item is damaged" and "no items are damaged" (let's denote the probability of this event as $Q_{1}$) are complementary, so $P_{1}+Q_{1}=1$. Therefore, the probability that at least one item will be damaged is

$$
P_{\mathrm{i}}=1-Q_{\mathrm{i}}=1-P_{500}(0)=1-\mathrm{e}^{-1}=1-0.36788=0.632
$$