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3 years ago

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Suppose there was a cancer diagnostic test was 95% accurate both on those that do and those do not have the disease. If 0.4% of

the population have cancer, compute the probability that a tested person has cancer given that his test result indicates.

1 answer:

Musya8 [376]3 years ago

7 0

Complete Question

Suppose there was a cancer diagnostic test was 95% accurate both on those that do and 90% on those do not have the disease. If 0.4% of the population have cancer, compute the probability that a particular individual has cancer, given that the test indicates he or she has cancer.

Answer:

The probability is $P(C | A) = 0.0042$

Step-by-step explanation:

From the question we are told that

The probability that the test was accurate given that the person has cancer is

$P(A | C) = 0.95$

The probability that the test was accurate given that the person do not have cancer is

$P(A | C') = 0.90$

The probability that a person has cancer is

$P(C) = 0.004$

Generally the probability that a person do not have cancer is

$P(C') = 1- P(C)$

=> $P(C') = 1- 0.004$

=> $P(C') = 0.996$

Generally the probability that a particular individual has cancer, given that the test indicates he or she has cancer is according to Bayes's theorem evaluated as

$P(C | A) = \frac{P(A | C) * P(C)}{P(A|C) * P(C) + P(A| C') * P(C')}$

=> $P(C | A) = \frac{ 0.95 * 0.004 }{ 0.95 * 0.004 + 0.90 * 0.996}$

=> $P(C | A) = 0.0042$

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Transition Matrix:

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Transition matrix P raised to the power 3 (at k = 3)

$P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

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$P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]$

Transition matrix P raised to the power 4 (at k = 4)

$P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]$

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$P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

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S = [ 0.2069 , 0.7931 ]

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