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Pie
3 years ago
8

A test for the presence of a certain disease has probability 0.2 of giving a false-positive result (indicating that an individua

l has the disease when this is not the case) and probability 0.1 of giving a false-negative result. Suppose that 10 individuals are tested, 5 of whom have the disease and 5 whom do not. Let X be the number of positive readings that result.
a. Explain why X does not have a binomial distribution.

b. Find the probability that exactly 3 of the 10 test results is positive?

Mathematics
1 answer:
drek231 [11]3 years ago
4 0
For the part b of the question please find the attached image.

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The probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.

Step-by-step explanation:

Let a set be events that have occurred be denoted as:

S = {A₁, A₂, A₃,..., Aₙ}

The Bayes' theorem states that the conditional probability of an event, say <em>A</em>ₙ given that another event, say <em>X</em> has already occurred is given by:

P(A_{n}|X)=\frac{P(X|A_{n})P(A_{n})}{\sum\limits^{n}_{i=1}{P(X|A_{i})P(A_{i})}}

The disease Breast cancer is being studied among women of age 60s.

Denote the events as follows:

<em>B</em> = a women in their 60s has breast cancer

+ = the mammograms detects the breast cancer

The information provided is:

P(B) = 0.0312\\P(+|B)=0.81\\P(+|B^{c})=0.92

Compute the value of P (B|+) using the Bayes' theorem as follows:

P(B|+)=\frac{P(+|B)P(B)}{P(+|B)P(B)+P(+|B^{c})P(B^{c})}

            =\frac{(0.81\times 0.0312)}{(0.81\times 0.0312)+(0.92\times (1-0.0312)}\\

            =\frac{0.025272}{0.025272+0.891296}

            =0.02757\\\approx0.0276

Thus, the probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.

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