Please explain to me how to solve this problem ---TRIAL QUESTIONS
1. A technician needs to inspect 5 different machines in a day. In how many different orders can the technician schedule the inspections?
2. A quality inspector selects 3 circuit boards from a batch of 7 for testing. How many different groups of 3 boards can be selected?
3. A diagnostic tool has a 95% detection rate. If disease prevalence is 2%, what concept is required to compute true detection rate in population?
4. Two pumps operate independently. The probability that Pump A fails is 0.2, and the probability that Pump B fails is 0.15. What is the probability that both pumps fail on the same day?
5. A device produces output X={0,1,3} with probabilities 0.5,0.3,0.2 respectively. Find E(X).
6. If P(Y=0)=0.3 and P(Y=1)=0.4, what must P(Y=2) be?
7. A pie chart is most effective for showing parts of a whole in categorical data.
8. The number of ways to arrange 5 books on a shelf is 5
9. If two events are independent, the probability of their intersection equals the product of their probabilities
10. The area under the probability distribution function of a continuous random variable is always 1
11. A shipment contains 4 defective and 6 non-defective microchips. If two microchips are selected at random without replacement, what is the probability that both are non-defective?
12. During a maintenance check, the probability that a machine overheats is 0.4, and the probability that the technician forgets a tool is 0.1. Assuming independence, what is the probability that both events occur?
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Let's explore two fundamental counting problems. First, we have a permutation problem where order matters - arranging 5 machines for inspection. Second, we have a combination problem where order doesn't matter - selecting 3 circuit boards from 7. These represent the two main types of counting problems in probability and statistics.
For the first problem, we need to find the number of ways to arrange 5 different machines. This is a permutation problem because the order of inspection matters. For the first position, we have 5 choices. For the second position, we have 4 remaining choices, and so on. The total number of arrangements is 5 factorial, which equals 5 times 4 times 3 times 2 times 1, giving us 120 different possible orders.
For the second problem, we need to select 3 circuit boards from 7, where order doesn't matter. This is a combination problem. We use the combination formula C of 7 choose 3, which equals 7 factorial divided by 3 factorial times 4 factorial. This simplifies to 7 times 6 times 5, divided by 3 times 2 times 1, which equals 210 divided by 6, giving us 35 different possible groups.
Now let's solve probability problems involving independent events. For problem 4, we have two pumps operating independently with failure probabilities of 0.2 and 0.15. For problem 12, we have a machine with overheat probability 0.4 and a technician with tool-forgetting probability 0.1. The key principle is that for independent events, the probability of both occurring equals the product of their individual probabilities. So we get 0.2 times 0.15 equals 0.03, and 0.4 times 0.1 equals 0.04.