ST107 Statistics Exercise 3 Bayes’ Total Probability Formula

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ST107 Exercise 3 In this exercise you will practise the total probability formula, Bayes’ formula and work with simple discrete probability distributions. Question 1 requires you to determine probabilities of events for a new random variable, T , derived from the original random variable, X. Question 2 requires the total probability formula in (a), and Bayes’ formula in (b). A probability tree (or tree diagram) is an alternative way to solve this problem. You are welcome to use this approach to check your answers to (a) and (b). However, you should try to use the total probability formula and Bayes’ formula for the purposes of this exercise! Question 3 requires you to work with the binomial and Poisson distributions, and you may find it convenient to use the result that P (A) = 1 − P (Ac ). Finally, Question 4 asks you to work with a binomial random variable, and a transformation of it. 1. A random variable X can take either the value 3 or the value 6. The respective probabilities of its taking these values are: P (X = 3) = 0.65 and P (X = 6) = 0.35. If twelve independent values of X are sampled, and the total of the twelve values is T , find the probability that T = 40 and the probability that T = 45. 2. The London Special Electronics company (LSE) is investigating a fault in its manufacturing plant. Tests done so far prove that the fault is at one of three locations – A, B or C. Taking all the initial tests into account, the company initially assesses the chances of the fault being at each site as follows: Suspect site Probability of it being the site of the fault A 0.6 B 0.3 C 0.1 The company now performs a new test which is expected to improve the identification of the correct site of the fault, but (like most tests) it is not entirely accurate. • If the fault is at A, then there is a 80% chance of a correct identification. That is, given that A is the site of the fault, then the probability that the new test says that the fault is at A (given the initial assessment that the probability is 0.6) is 0.8, and in this case the probabilities of either possible error (the new test saying the fault is at B or C, again given the initial assessment) are equal. • If the fault is at B, then there is a 70% chance of a correct identification, and in this case too the probabilities of either possible error are equal. • If the fault is at C, then there is an 60% chance of a correct identification, and in this case too the probabilities of either possible error are equal. 1 Use probability formulae to answer the following. (a) What is the probability that the new test will (rightly or wrongly) identify A as the site of the fault? (b) If the new test does identify A as the site of the fault, find: i. the company’s revised probability that C is the site of the fault ii. the company’s revised probability that B is not the site of the fault. 3. Suppose it is known that 15% of a large population believe cryptocurrencies should be a widely accepted form of payment by retailers. It is proposed to conduct a survey of these people as part of an effort to encourage retailers to introduce this payment method. (a) Find the probability that, in a pilot survey of 16 people, at least three of them favour cryptocurrencies as a payment method. (b) Using a suitable approximation, find the probability of at least three people in favour in a larger random sample of 50 people. In both parts, explain any assumptions you are making in order to justify the methods of calculation which you use. 4. A random variable X has a binomial distribution with n = 6 and π = 0.4. (a) Write out the probability distributions of X and (X − 2)2 . (b) Find the mean and the variance of X, and the mean of (X − 2)2 using: i. the distributions derived in (a) ii. the properties of the expectation operator, E(·), and noting that the mean and variance of a binomial random variable are n π and n π (1 − π), respectively. 2 ...
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Tutor Answer

School: New York University

it's 4 hours now..I'll just send out the solution, although there is no assurance that you'll make this appear that I sent it overdue :(
Here, I added additional equations / details :)


The probability that T = 40 is zero. This is because there is no combination X = 3 and X = 6 that
would result to a value of 40. This is because 40 is not a factor of either 3 or 6. For the case of T
= 45, the only possible combination of 3s and 6s for the sum to be 45 is as follows:
X = 3 : 9 values
X = 6 : 3 values
The probability of this happening follows a binomial distribution as follows:
P(x) = nCx (p)x (1 – p)n – x
n = number of samples
x = value of the random variable
p = pro...

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Tutor went the extra mile to help me with this essay. Citations were a bit shaky but I appreciated how well he handled APA styles and how ok he was to change them even though I didnt specify. Got a B+ which is believable and acceptable.

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