Please help.... fast

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 are the next 3 terms in the pattern -24,-14,-4,6...

zmey [24]

The pattern is plus 10 for each time

Formula recursive
A_n = A_n -1 + 10

So the next term is 5th

A_5 = A_5 -1 + 10

A_5-1 = A_4 +10

5th = 6+10 = 16

I think you can do it for the next 2 terms.

4 0

3 years ago

Read 2 more answers

B. Translate each word into algebraic eqaution. (Use any letter)

Andrei [34K]

Twice a number is the same as
A number used twice
If that number where represent by x
Then
If x was used twice would look like
What

4 0

2 years ago

If your heart beats 80 times per minute and you live to be 90 years old, estimate the number of times your heart beats during yo

umka2103 [35]

The heart beat is given in units of beats per minute, therefore before we can do any calculations let us make sure that all units are similar by converting the age in terms of minutes.

We know that there are 365 days per year, therefore:

90 years (365 days / year) = 32,850 days

We know that in a day, there are 24 hours, therefore:

32,850 days (24 hours / day) = 788,400 hours

And in an hour, there are 60 minutes, therefore:

788,400 hours (60 minutes / hour) = 47,304,000 minutes

Now calculating for the total heart beats in a lifetime:

total heartbeats = (80 beats / minute) * 47,304,000 minutes

total heartbeats = 3,784,320,000

Writing this into scientific notation:

3.78 x 10^9 beats

or

<span>3.8 x 10^9 beats</span>

8 0

3 years ago

Frank bought 6 pens and 3 highlighters in the school store and paid $21 for the whole purchase. His friend, Andy, paid $19 for 2

vovangra [49]

Answer:one pen = $1.50

Step-by-step explanation:

See picture for the math if that helps

Using p for pens, h for highlighter,

A for first equation

B for second equation

I used equation A to reduce to a usable substitute by dividing all by 3

And then isolating “h”.

Now substitute the value of “h” into equation B and solve for “p”

P=$1.50

To check your work, substitute p ($1.50) into equation B and solve for “h”

h=$4.00

Now substitute the p and h values into either equation A or B and solve, they equal the numbers given

5 0

2 years ago

Please help.... fast​

Please help.... fast