A batch of 140 semiconductor chips is inspected by choosing a sample o

Step 1
Given:
${\displaystyle n=140}$
${\displaystyle r=5}$
10 of the 140 chips are non-conforming.
Definition permutation (order is important):
${\displaystyle P_{n,r}=\frac{n!}{(n-r)!}}$
Definition combination (order is not important):
${\displaystyle C_{n,r}=\frac{n!}{r!(n-r)!}}$
with ${\displaystyle n\ne n\cdot (n-1)\cdot \cdots \cdot 2\cdot 1}$
Step 2
a. Order is not important, because it is not important whether a chip was selected first or was selected last and thus we need to use the definition of a combination.
${\displaystyle C_{140,5}=\frac{140!}{5!(140-5)!}=\frac{140!}{5!135!}=416,965,528}$
Step 3
b. The sample contains 1 nonconforming chip, which means that 1 nonconforming chip was selected from the 10 nonconforming chips:
${\displaystyle C_{10,1}=\frac{10!}{1!(10-1)!}=\frac{10!}{1!9!}=10}$
The sample contains 1 nonconforming chip out of 5 chips and thus also contains 4 conforming chips, which means that 4 conforming chip was selected from the 130 conforming chips (140 chips in total of which 10 are nonconforming, thus 130 have to be conforming):
${\displaystyle C_{130,4}=\frac{130!}{4!(130-4)!}=\frac{130!}{4!126!}=11,358,880}$
The total number of samples in which to select a sample with 1 nonconfirming chip is then the product of the number of ways to select 1 nonconforming chip and the number of ways to select 4 conforming chips.
${\displaystyle 10\times 11,358,880=113,588,800}$
Step 4
c. Let us first determine the number of samples that contain no nonconforming chips. The number of samples with at least one nonconforming chips is then the total number of samples decreased by the number of samples with no conforming chips.
The sample contains 0 nonconforming chips, which means that 0 nonconforming chips was selected from the 10 nonconforming chips:
${\displaystyle C_{10,0}=\frac{10!}{0!(10-0)!}=\frac{10!}{0!10!}=1}$
The sample contains 0 nonconforming chips out of 5 chips and thus also contains 5 conforming chips, which means that 5 conforming chip was selected from the 130 conforming chips (140 chips in total of which 10 are nonconforming, thus 130 have to be conforming):
${\displaystyle C_{130,4}=\frac{130!}{5!(130-5)!}=\frac{130!}{5!125!}=286,243,776}$
The total number of samples in which to select a sample with 0 nonconforming chips is then the product of the number of ways to select 0 nonconforming chips and the number of ways to select 5 conforming chips.
${\displaystyle 1\times 286,243,776=286,243,776}$
The number of samples with at least one nonconforming chips is then the total number of samples (part a) decreased by the number of samples with no conforming chips.
${\displaystyle 416,965,528-286,243,776=130,721,752}$

a) The lot size is 140, the sample size is 5
$$(\begin{array}{c}140\\ 5\end{array})=146,965,528$$
b) How many samples contain 4 good and i bad chip:
$$(\begin{array}{c}130\\ 4\end{array})(\begin{array}{c}10\\ 1\end{array})=113,588,800$$
c) How many samples contain:
(4 good and 1 bad) or (3 good and 2 bad) or (2 good and 3 bad) or (1 good and 4 bad) or (0 good and 5 bad)
$$=(\begin{array}{c}130\\ 4\end{array})(\begin{array}{c}10\\ 1\end{array})+(\begin{array}{c}130\\ 3\end{array})(\begin{array}{c}10\\ 2\end{array})+(\begin{array}{c}130\\ 2\end{array})(\begin{array}{c}10\\ 3\end{array})+(\begin{array}{c}150\\ 1\end{array})(\begin{array}{c}10\\ 4\end{array})+(\begin{array}{c}130\\ 0\end{array})(\begin{array}{c}10\\ 5\end{array})$$
${\displaystyle =113,588,800+16,099,200+1,006,200+27,300+252}$
${\displaystyle =130,721,752}$

The velocity vector addition:
$$\begin{gathered} {\overrightarrow{v}}_{p/a}+{\overrightarrow{v}}_{a/g}={\overrightarrow{v}}_{p/g} \\ a)\sin \theta =\frac{|{\overrightarrow{v}}_{a/g}|}{|{\overrightarrow{v}}_{p/a}|}=\frac{40km/h}{320km/h}=0.25 \\ \theta ={14.5}^{\circ },\text{north of west} \\ b)|{\overrightarrow{v}}_{p/g}|=\sqrt{(|{\overrightarrow{v}}_{p/a}|{)}^2-(|{\overrightarrow{v}}_{a/g}|{)}^2}=\sqrt{(320km/h{)}^2-(80km/h{)}^2}=80\sqrt{15}km/h\approx 310km/h \end{gathered}$$