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THE PROBLEMS, AND BACKGROUND INFO

Hi, I was checking the answers of this homework and was wondering why in problem C, my teacher writes NEEDS 3; and she does not say that in any of the above. I believe the answer should be 81/1000, but I just want to hear what you think. Thanks!

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    $81/1000$ is the probability that the person is late only on one *specific* day, say Monday. You want the probability that the person is late only on Monday *or* only on Tuesday *or* only on Wednesday. Each of these probabilities is the same, so you can just compute one and multiply it by three. – 2017-01-13

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Your teacher is right. There are three possibilities. He could be late Monday, Tuesday, or Weds. Each of these has probability 81/1000 of happening, so the total probability is three times that. Notice in problem B they specify exactly which days he is late, so you don't need any extra factor.

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    Thanks. I wonder why if I want to get 10 days, where one of them he is late theoretically, I would have this equation: `(81/10)*(1/10)` and it doesn't add up to 100%. I was wondering the logic behind this, and why it won't multiply to 100% and what you would have to do to make it. – 2017-01-13
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    @Blake I don't understand your question, sorry. Could you write it out in more detail? – 2017-01-13
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    I just edited it, sorry for being confusing ;) I wanted to show "What is the probability of him being late once in 10 days?", but realized you would do 1/10 * 10 and get 1, or 100% ;) – 2017-01-13
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    @Blake "what is the probability of him being late exactly once in 10 days" would be $\binom{10}{1}\cdot 0.1^1\cdot 0.9^{9}$ which is not equal to one... The **expected number of days** that he is late in ten days will be one, but that is different than the probability that he is late one day in ten. – 2017-01-13
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    That is what I wanted to know. Thanks a bunch :) – 2017-01-13
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    @Blake as an aside, what is the probability he will be late once in twenty days is clearly not going to be equal to $2$ since there is no such thing as a probability greater than $100\%$ (in conventional probability) – 2017-01-13
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    I knew that, thank you though. Although,I believe it would never be 100%, it would limit at 99.99..., etc, at such a high number because unless it is 1/1 chance. Thanks for all the help! – 2017-01-13