Example 7. Quality control has established that, on average, 75 out of every 100 parts are not defective. Construct the binomial probability distribution for the number of suitable parts out of 6 parts taken at random.
Solution. From the condition of the problem, it follows that $p=0.75, q=0.25, n=6$. According to the Bernoulli formula, we find:

$$
\begin{aligned}
& P_{6}(0)=1 \cdot(0.25)^{6} \approx 0.0002 ; \quad P_{6}(1)=6 \cdot(0.75) \cdot(0.25)^{5} \approx 0.004 ; \\
& P_{6}(2)=15 \cdot(0.75)^{2} \cdot(0.25)^{4} \approx 0.033 ; \quad P_{6}(3)=20 \cdot(0.75)^{3} \cdot(0.25)^{3} \approx 0.132 \\
& P_{6}(4)=15 \cdot(0.75)^{4} \cdot(0.25)^{2} \approx 0.297 ; \quad P_{6}(5)=6 \cdot(0.75)^{5} \cdot(0.25) \approx 0.356 \\
& P_{6}(6)=1 \cdot(0.75)^{6} \approx 0.178
\end{aligned}
$$

The distribution law of the random variable $X$ - "the number of standard parts out of 6 taken at random" can be given by the following table:

| $X$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.000 | 0.004 | 0.033 | 0.132 | 0.297 | 0.356 | 0.178 |

Let's verify that the equality (3.2.4) holds, i.e., the sum of all probabilities is equal to one:

$$
0.004+0.033+0.132+0.297+0.356+0.178=1
$$

The graphical representation of this binomial distribution is given in Fig. 3.1.

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-189.jpg?height=574&width=816&top_left_y=24&top_left_x=173)

## Fig. 3.1