1. 3 men and 10 women come to a dance class. What is the number of different Constellations of 3 dance couples each (a woman and a man)? --> 604800
2. 13 people from a village want to carpool to the nearest town. There are two Vehicles available with space for (11) and (5) people. How many options are there To split 13 people between the two vehicles? --> 1287
3. In a delivery of 60 devices are 23 defective devices. The devices are through theirs Device number distinguishable, but it is not clear which is the defective one without closer examination. What is the probability that there are 4 defective devices in a random sample of 9 devices? --> 26,11 %
4. A test for the detection of the coronavirus brings in 99.4% of all subjects infected with the coronavirus a positive test result. It gives a negative test result in 94% of all non-infected subjects. They have the test done at a time when 0.38% of the population is acutely infected with the coronavirus. Assuming the test gives a positive result, what is the probability that You are actually infected with the coronavirus? --> 5,944 %
5. A test for the detection of a different virus is available at a different time than in task 4. At this point in time, the test has the following characteristics: The test gives a positive result in 98% of all test subjects infected with the virus; The test yields a negative test result in 93.4% of all non-infected test subjects; The likelihood that a person who tests positive is actually infected with the virus, is 89.6%. What percentage of the population is infected with the virus at this point in time? -->?
6. As an avid archer, you have reached a level where you can hit a target 50 m away with a probability of 16% in the innermost ring of the disc. Calculate the probability that you will hit the innermost ring at most three times with 23 shots. -->48,59 %
7. There are 76 winning tickets in a lottery drum with 1000 tickets. You buy one lot at a time, like that long until you make a profit (then do not buy any more tickets). Calculate the exact probability that you will have to buy 5 tickets. Don't do any Approximation when calculating the probability by. --> 7,0656 %
8. There are 76 winning tickets in a lottery drum with 1000 tickets. You buy one lot at a time, like that long until you make a profit (then do not buy any more tickets). Calculate the approximate probability that you will have to buy more than 23 lots. The approximation to use is that you do the pull with pulling with replacement. --> 16,2351 %
9. The lottery drum with the 1000 tickets is considered again, among which there are 76 winning tickets. The Prizes for the 76 winning tickets are distributed as follows: 3 lots bring the main prize of 1000 euros 10 tickets bring a profit of 200 euros 63 tickets are associated with a profit of 10 euros The remaining 924 lots are rivets, so they have a profit of 0 euros. Calculate the expected profit in euros if you only buy one ticket (namely the first ticket) from the pulling the lottery wheel. -->8.97?

See a short discussion of the solutions below. In some cases I get the same number as you, just rounded differently. If you need additional decimal precision or if some part is not clear to you don't hesitate to ask.

1. The correct number is ${3\choose 3}\cdot {10\choose 3}\cdot 3!=720$ (choose $3$ from the $3$ men, $3$ from the $10$ women and then choose a way to pair the chosen $3$ men with the chosen $3$ women).
2. The correct number is ${13\choose 2}+{13\choose 3}+{13\choose 4}+{13\choose 5}= 2366$ (there are ${13\choose 2}$ many configurations where $2$ people goe in the second car and $11$ in the first, then ${13\choose 3}$ configurations where $3$ people go in the second car and $10$ in the first, etc)
3. The answer depends on what is meant with "4 defective devices in a sample of 9", this could be understood as "exactly 4 from 9 are defective" or "atleast 4 from 9 are defective". In the first case the probability is $$\frac{{23\choose 4}\cdot {37\choose 5}}{{60\choose 9}}\approx 0.2611$$ ie $26.11\%$. In the second case you get:
$$\sum_{k=4}^9\frac{{23\choose k}\cdot {37\choose 9-k}}{60\choose 9}\approx 0.477$$ ie $47.7\%$.
4. Let $I$ denote the event that you are infected, and $T$ the event that the test is positive. Then
$$P(I) = 0.0038,\quad P(T\mid I )= 0.994,\quad P(\neg T\mid \neg I) =0.94$$ we want to find $P(I\mid T)$. Using Bayes' theorem: $$P(I\mid T) = \frac{P(T\mid I) P(I)}{P(T)}$$ the number $P(T)$ is still unknown, but here we use the formula $$P(T) = P(T\mid I) P(I) + P(T\mid \neg I) P(\neg I) = P(T\mid I) P(I) +(1-P(\neg T\mid I)) (1-P(I))$$ in numberes then: $$P(T) = 0.994\cdot 0.0038+(1-0.94)\cdot (1-0.0038) \approx 0.06355$$ and then $$P(I\mid T) = \frac{0.994\cdot 0.0038}{0.06355}\approx 0.0594$$ so about $5.94\%$.
5. Using the same labels we now read off: $$P(T\mid I) = 0.98, \quad P(\neg T\mid \neg I)= 0.934, \quad P(I\mid T) = 0.896$$we want to find  $P(I)$, we use the formula: $$P(I) = \frac{P(I\mid T) P(T)}{P(T\mid I)}$$ Here we still need to find $P(T)$. Note that$$P(T) = P(T\mid I ) P(I) + P(T\mid \neg I) P(\neg I) = P(T\mid I) P(I) + (1-P(\neg T\mid\neg I)(1-P(I))$$ so $$P(T) = P(I) \cdot (P(T\mid I) + P(\neg T\mid \neg I) -1) + 1-P(\neg T\mid \neg I)= 0.914\cdot P(I) +0.076$$
We put this into our equation for $P(I)$ and get
$$P(I) = \frac{0.896\cdot (0.914 \cdot P(I)+0.076)}{0.98}\approx 0.8357\cdot P(I) +0.0695$$
so $P(I) \approx \frac{0.0695}{1-0.8357}\approx 0.423$, ie approximately $42.3\%$.
6. The probability is $$\sum_{k=0}^3 {23\choose k}0.16^k\cdot (1-0.16)^{23-k} \approx 0.486$$ or approximately $48.6\%$.
7. The probability is $\frac{(924!/920!)\cdot 76}{(1000!/995!)}\approx 0.0556$ or about $5.56\%$.
8. You need to buy more than $23$ lots if you draw $23$ failures in a row. Since we are approxmating this by using replacement we get:$$(\frac{924}{1000})^{23}\approx 0.1624$$ as a probablity, which is $16.24\%$.
9. The expected winnings of one ticket is:
$$1000\cdot \frac{3}{1000}+ 200\cdot \frac{10}{1000}+10\cdot \frac{63}{1000}+0\cdot \frac{924}{1000} = 5.63$$
note that to find the expected "profit" we should subtract the cost of a ticket from the expected winnings. But since no such cost is given I assume the word "profit" was just poorly chosen.
• Feivel12

Thanks a lot for your effort.