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Here is a guide to the questions on the exam:
#1: rules of probability (Sections 4.3 & 4.5, in particular the Addition Rule and Multiplication Rule)
#2 & #4: discrete random variables, probability distributions, expected value, variance/standard deviation (Sections 4.2-4.4)
#3 & #6: binomial random variables (Sec 4.5)
#6: normal random variables (Sec 6.3)
If you get stuck, look at the examples and homework exercises for the corresponding sections. You can ask me questions about examples in the book or homework exercises.
You might find looking at my Google spreadsheet useful for the exam. I’ve included solutions for some of the homework exercises.
In particular, the “HW7 – Binomial” tab has solutions for a couple of HW exercises from Sec 4.5, including using =binomdist(i,n,p,true) to calculate cumulative binomial probabilities–which could be useful for #3 & #6 on the exam
You can view the Google spreadsheet here: https://docs.google.com/spreadsheets/d/1pgVX90vBiFrJgdiTKfqIv23Kv_eJC_v0lwD89upj2Rk/edit?usp=sharing
Hello Professor, i’m confused in q#2(a). there is only one blue marble, if two marble drawn without replacement then there won’t be any marble left for second time..i was wondering if Y would be 0. cause it would be like (1/10)X (0/9) which will give us 0… i’m just confused if i’m getting it right..
You’re on the right track. The calculation you did shows that the probability of drawing two blue marbles is 0–that makes sense since as you said there’s only one blue marble to start, so it’s impossible to draw two. But that doesn’t mean the only possible value of Y is 0. It shows that 2 is not a possible value of Y.
Think about the other possibilities for you might draw (hint: look at part (b)), and consider what the value of Y is in each of those cases.
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