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# need help with some stats questions watch

1. hi, i really need help with a good few questions over the next few weeks. 3 of the most important can be found below. any help will be greatly appreciated (also with rep)

7. A new test has been devised for dectecting a particular type of cancer. If the test is applied to a person who has this type of cancer, the probability that the person will have a positive reaction in the test is 96%. If the test is applied to a person who does not have this type of cancer, the probability that the person will have a positive reaction is 1%. Suppose that in the general population, one person out of every 100,000 people has this type of cancer. If a person selected at random has a positive reaction to the test, what is the probability that they have this type of cancer.

8. (a) What is a Bernoulli trial?
(b) Suppose a fair coin is tossed until at least one head and at least one tail have been obtained. By considering the different outcomes of the first toss, determine the probability that two tosses will be required, and that three tosses will be required. Can you generalise this to the probability that n>=2 tosses will be required?

9.(a) In a linear sample space (the real numbers) there is a probability of 0.0013 for the outcome to be in the interval (-0.0005, 0.0005)
(i) assuming that the probability density function is constant over this small interval, what is its value there?
(ii) comment on the value obtained.
(b)There is a constant probabilty density of 1.1 over the interval (0,0.3) and over the interval (0.3,0.4) there is a constant probability density of 1.3. What is the probability for the outcome to be in the interval (0,0.4)?

Thanks
2. (7)
Define two events: A = {Person has the type of cancer}, B = {Person's test is positive}. Then

P(A | B)
= P(A and B) / P(B)
= 0.00001*0.96 / (0.00001*0.96 + 0.99999*0.01)
= 0.000959

(so the test is almost useless!)

(8)
(a)
A Bernoulli trial is a random variable that can take only two values.

(b)
Let X be the number of tosses required.

If the first toss is H then: X = 2 iff the second toss is T, which happens with probability 1/2.

If the second toss is T then: X = 2 iff the second toss is H, which happens with probability 1/2.

So P(X = 2) = 1/2.

--

If the first toss is H then: X >= 4 iff the second and third tosses are both H, which happens with probability 1/4.

If the first toss is T then: X >= 4 iff the second and third tosses are both T, which happens with probability 1/4.

So

P(X >= 4) = 1/4

P(X = 3)
= P(X >= 3) - P(X >= 4)
= 1/2 - 1/4
= 1/4

--

If the first toss is H then: X >= n + 1 iff the next (n - 1) tosses are all H, which happens with probability (1/2)^(n - 1).

If the first toss is T then: X >= n + 1 iff the next (n - 1) tosses are all T, which happens with probability (1/2)^(n - 1).

So P(X >= n + 1) = (1/2)^(n - 1).

--

P(X = n)
= P(X >= n) - P(X >= n + 1)
= (1/2)^(n - 2) - (1/2)^(n - 1)
= (1/2)^(n - 1)

(9)
(a)
(i)
0.0013/0.001 = 1.3

(ii)
The answer to (i) illustrates that, while probabilities are always between 0 and 1, PDFs can take values above 1.

(b)
1.1*0.3 + 1.3*0.1 = 0.46

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