Answer:
Amy should not use her previous luck of rolling a 3 four times because theoretical probability remains the same even after an event favoring one side has occurred by chance.
Step-by-step explanation:
Assuming that this number cube has 6 sides, there is only 1 side that has a 3 on it. For this reason, every time the cube is rolled there is a 1 in 6 chance that the number will be a 3. If she rolls the number cube 240 times, 1 out of 6 rolls would be a 3. This means that 240/6 equals the number of times she would get a 3. Moreover, Amy would role the number 3 on 40 different times after rolling the cube for a total of 240.
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Answer:
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by

In which r is the number of failures we want and p is the probability of a failure.
In this problem, we have that:
r = 1 because we want the first failed unit.
![p = 0.4[\tex]So[tex]E = \frac{r}{p} = \frac{1}{0.4} = 2.5](https://tex.z-dn.net/?f=p%20%3D%200.4%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3ESo%3C%2Fp%3E%3Cp%3E%5Btex%5DE%20%3D%20%5Cfrac%7Br%7D%7Bp%7D%20%3D%20%5Cfrac%7B1%7D%7B0.4%7D%20%3D%202.5)
The expected number of systems inspected until the first failed unit is 2.5
I think 4 and 6 lcm is 12