91影视

An unordered subset is referred to as a combination. The number of combinations of size \({\rm{k}}\)for \({\rm{n}}\) persons in a group is indicated as \({{\rm{C}}_{{\rm{k,n}}}}\)

\(\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\)

This we interpret as \({\rm{k}}\)items from \({\rm{n}}\)elements. The proposal is this:

\({{\rm{C}}_{{\rm{k,n}}}}{\rm{ = }}\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right){\rm{ = }}\frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\)

Because we don't care about the order, we may utilize combinations.

\({{\rm{C}}_{{\rm{5,25}}}}{\rm{ = }}\frac{{{\rm{25!}}}}{{{\rm{5!(25 - 5)!}}}}{\rm{ = }}\frac{{{\rm{25!}}}}{{{\rm{5!20!}}}}{\rm{ = 53130}}\)

Out of 25, there are five ways to take five keyboards for a complete check.

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\)ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

We want to know how many options we have for the first element.

We can accomplish it since there are just two electric flaws out of a total of six.

For the first element, we want to know how many different ways we may choose two electric flaws from a total of six. That is something we can accomplish in

\({{\rm{C}}_{{\rm{2,6}}}}{\rm{ = }}\frac{{{\rm{6!}}}}{{{\rm{2!(6 - 2)!}}}}{\rm{ = }}\frac{{{\rm{6!}}}}{{{\rm{2!4!}}}}{\rm{ = 15}}\)

ways. And how many different ways can we extract \({\rm{3(5 - 2 = }}\)for the second element \({\rm{3}}\), we took 2) keyboards from the remaining \({\rm{19(25 - 6 = 19}}\)non-electric) keyboards. That is something we can accomplish in

\({{\rm{C}}_{{\rm{3,19}}}}{\rm{ = }}\frac{{{\rm{19!}}}}{{{\rm{3!(19 - 3)!}}}}{\rm{ = }}\frac{{{\rm{19!}}}}{{{\rm{3!16!}}}}{\rm{ = 969}}\)

ways. As a result, the final response is

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{ = 15*969 = 14535}}\)

The union of two events, \({{\rm{A}}_{\rm{1}}}\)- "exactly \({\rm{4}}\) keyboards have a mechanical flaw" and \({{\rm{A}}_{\rm{2}}}\)- "exactly \({\rm{5}}\) keyboards have a mechanical defect," is Event A - "at least \({\rm{4}}\) keyboards will have a mechanical defect." We use a formula to compute probability.

Therefore

\(\begin{aligned}P(A)&=P\left( {{A}_{1}}\grave{E}{{A}_{2}} \right)=P\left( {{A}_{1}} \right)+P\left( {{A}_{2}} \right)= \\&=\frac{\left( \begin{aligned}19 \\4 \\\end{aligned} \right)\left( \begin{aligned}6 \\1 \\\end{aligned} \right)}{\left( \begin{aligned}25 \\5 \\\end{aligned} \right)}+\frac{\left( \begin{aligned}19 \\5 \\\end{aligned} \right)\left( \begin{aligned}6 \\0 \\\end{aligned} \right)}{\left( \begin{aligned}25 \\5 \\\end{aligned} \right)} \\&=\frac{3876*6}{53130}+\frac{11628*1}{53130} \\&=0.438+0.219 \\&=0.657, \\\end{aligned}\)

(1): we leverage the fact that the events are discontinuous in this case;

(2): similarly to \({\rm{(b)}}\), we calculate the number of favorable outcomes in \({{\rm{A}}_{\rm{1}}}\) and \({{\rm{A}}_{\rm{2}}}\) and the number of outcomes in the sample space is given in \({\rm{(a)}}\)