What do you subtract from each side

Can somebody simply these matrixs for me please I will give out brainlist to first correct answer

shtirl [24]

I think is D I am so sorry if I’m wrong

8 0

3 years ago

It is known that there only is 1% chance of getting a disease. a test is being devised to detect the disease. the probability th

Cerrena [4.2K]

Suppose $D$ is the event that a given patient has the disease, and $P$ is the event of a positive test result.

We're given that

$\mathbb P(D)=0.01$
$\mathbb P(P\mid D)=0.98$
$\mathbb P(P^C\mid D^C)=0.95$

where $A^C$ denotes the complement of an event $A$ .

a. We want to find $\mathbb P(P^C)$ . By the law of total probability, we have

$\mathbb P(P^C)=\mathbb P(P^C\cap D)+\mathbb P(P^C\cap D^C)$

That is, in order for $P^C$ to occur, it must be the case that either $D$ also occurs, or $D^C$ does. Then from the definition of conditional probability we expand this as

$\mathbb P(P^C)=\mathbb P(D)\mathbb P(P^C\mid D)+\mathbb P(D^C)\mathbb P(P^C\mid D^C)$

so we get

$\mathbb P(P^C)=0.01\cdot0.02+0.99\cdot0.95=0.9407$

b. We want to find $\mathbb P(D\mid P)$ . Now, we can use Bayes' rule, but if you're like me and you find the formula a bit harder to remember, we can easily derive it.

By the definition of conditional probability,

$\mathbb P(D\mid P)=\dfrac{\mathbb P(D\cap P)}{\mathbb P(P)}$

We have the probabilities of $P$ / $P^C$ occurring given that $D$ / $D^C$ occurs, but not vice versa. However, we can expand the probability in the numerator to get a probability in terms of $P$ being conditioned on $D$ :

$\mathbb P(D\cap P)=\mathbb P(D)\mathbb P(P\mid D)$

Meanwhile, the law of total probability lets us rewrite the denominator as

$\mathbb P(P)=\mathbb P(P\cap D)+\mathbb P(P\cap D^C)$

or in terms of conditional probabilities,

$\mathbb P(P)=\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)$

so that

$\mathbb P(D\mid P)=\dfrac{\mathbb P(D)\mathbb P(P\mid D)}{\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)}$

which is exactly what Bayes' rule states. So we get

$\mathbb P(D\mid P)=\dfrac{0.01\cdot0.98}{0.01\cdot0.98+0.99\cdot0.05}\approx0.1653$

6 0

3 years ago

What is the value of n? N–2=10n+42 Enter your answer in the box. N =

Oliga [24]

$\frac{44}{9}$