Probability
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I only need help with problems 1, 2, and 3. Please see the following website for the complete problems:http://www.isye.gatech.edu/people/faculty/Robert_Foley/classes/2027/hmwk3.pdf
1. Suppose that the sample space S = {1, 2, 3, ...}. Let pk = Pr({k}) for k 2 S. In each of the following
cases, compute c. (a) Suppose that pk = c(5/6)k for k 2 S; (b) Suppose that pk = c(5/6)k/(k)! for
k 2 S.
2. Suppose that the sample space is S = [0,1). Let Bt = [t,1) for any t 0. Suppose that Pr(Bt) =
ce−6t for t 0. Compute (a) c, (b) Pr(B2), and (c) Pr([1, 2)).
3. Suppose we are dealt 6 cards from a standard well-shuffled deck. What is the probability that there
are (a) a six card flush? (b) 4 of one kind and 2 of another? (c) two triples? (d) 3 pairs? (e) 2 pairs?
(f) 1 pair? (g) at least one pair? You may leave your answer in terms of
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.
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Solution Summary
The probability sample space functions are examined. The standard well-shuffled deck is determined.
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1. Solution. Let us denote being head by H and being the number by N respectively when flipping one coin. We know that rolling one die corresponds to possible six results:1,2,3,4,5 or 6. So
a) A reasonable sample space S for this experiment is as follows.
S={{H,1},{H,2},{H,3},{H,4},{H,5},{H,6},{N,1},{N,2},{N,3},{N,4},{N,5},{N,6}}
Note: one sample point in S, say {H,1} means the result of the experiment is with the head of coin up and roll a 1.
b) {H,1} means the result of the experiment is with the head of coin up and roll a 1;
{H,2} means the result of the experiment is with the head of coin up and roll a 2;
{H,3} means the result of the experiment is with the head of coin up and roll a 3;
{H,4} means the result of the experiment is with the head of coin up and roll a 4;
{H,5} means the result of the experiment is with the head of coin up and roll a ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"
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