Some problems of \"Discrete Probability Theory\". Problem 1 is about Mutual inde

Question

Some problems of "Discrete Probability Theory".

Problem 1 is about Mutual independence:

Problem 2 is about Independence and correlation fucntions:

Problem 3 is about some probability calculations:

Thank you for your helping!!

We defined the concept of pairwise independence in this chapter. There is a related concept called mutual independence. Consider the set of random variables They are said to be mutually independent if for any subset of {AjJ containing n of these random variables, the marginal distribution satisfies the condition Obviously, mutual independence implies pairwise independence, The question is whether pairwise independence implies mutual independence Provide a proof or a counter-example.

Explanation / Answer

Question 1. Pairwise dependence does not implies mutual dependence.

Example - You toss a fair coin twice. The possible events are:

A. Both tosses give the same outcome (HH or TT). Probability = 1/4+1/4 = 1/2

B. The first toss is a heads (HT HH) = Probability = 1/4+1/4 = 1/2

C. The second toss is heads (TH HH) = Probability = 1/4+1/4 = 1/2

P(1) * P(2) * P(3) = 1/2 * 1/2 * 1/2 = 1/8.

P(1 2) = P(1 3) = P(2 3) = 1/4

But P(1 2 3)  = 1/8

P(1 2 3) is not equal to P(1) * P(2) * P(3)

Question 3

1.

P(no 3 in 10 rolls) = (5/6)^10 = 0.1615

2. P(atleast one 6 on 4 rolls) > P( at least one double 6 on 24 rolls)  

P(atleast one 6 on 4 rolls) = 1-(5/6)^4 = 0.5177

P( at least one double 6 on 24 rolls)  

We calculate the probability of rolling at least one double-six in 24 rolls of two dice. The probability we roll a double-six is, as you point out, 1/36.

So, on any roll, the probability of not getting a double six is 35/36.

The probability of "failure" 24 times in a row is therefore (35/36)^24.

So the probability of at least one "success" in 2424 rolls is 1(35/36)^24 = 0.4914

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