# Normal Distribution Calculating Probability

I am struggling with the following question:

A company which produces $1L$ beverages adjusts their machines in a way that the filling quantity is normally distributed. The mean is $\mu=995\,\text{cm}^3$ and the standard deviation is $\sigma = 5\,\text{cm}^3$.

To prevent manipulations, authorities take samples. If in a sample of five cans at least four cans contain more than $997\,\text{cm}^3$ and the fifth contains more than $995 \,\text{cm}^3$, nothing will be queried. With what probability are those manipulations discovered (1) in one sample (2) in 10 samples.

I tried the following:

$$\mu=995cm^3 \space\space\space\space \sigma=5cm^3$$

$$P(X>997cm^3)=1-\int_{-\infty}^{997}\frac{1}{{\sigma \sqrt {2\pi } }}e^{ - \frac{(x - \mu)^2}{2\sigma ^2}}\,\mathrm dx$$

$$P(X>995cm^3)=1-\int_{-\infty}^{995}\frac{1}{{\sigma \sqrt {2\pi } }}e^{ - \frac{(x - \mu)^2}{2\sigma ^2}}\,\mathrm dx$$

(1) $P=P(X>997cm^3)^4*P(X>995cm^3)$ since four have to contain more than $997 cm^3$ and one has to contain more than $995 cm^3$.

(2) $P=10*(1)$ since we have ten times more tries now.

I get for (1) $0.7%$ and for (2) $7%$ but according to the solutions' book I should get a percentage of about $98$ for (1) and $~100%$ for (2).

In (2), what you are looking for is the complement to no queries in all samples. Hence, if $p_{(1)}$ is the probability that nothing will be queried in one sample, you want $1-\left[p_{(1)}\right]^{10}$. If you just multiply it by 10 you could very well end up with a 'probability' which is larger than 1.