Which of the following is a valid counterexample to disprove

Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

Formula: $P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}$

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: $P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}$

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0

3 years ago

If a= 5, b= 8, and c= ½<br><br> 6a - 2b + bc

Rainbow [258]

Answer:

18

Step-by-step explanation:

6x5=30

2x8=16

8x1/2=4

30-16=14

14+4=18

8 0

3 years ago

In the Treasure, Treasure! video game, players gain and lose points by finding and losing treasure. In Level 2, Leslie earned

USPshnik [31]

Answer:

. finding and losing treasure. In level 2, Leslie earned -22 points when a storm blew a treasure chest off her ship. Then, she earned 34 ... a storm blew a treasure chest off her ship. Then, she earned 34 points for finding more treasure on the deserted island. What is the overall change in Leslie's score during level 2? ansver.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

R varies inversely as v2 if r=120 when v=1 ,find r when v=10

makvit [3.9K]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ r = k/v^2

Substitute the values in:

120 = k/1^2

Multiply both sides by 1^2 to find k

k = 120

Substitute new values in:

r = 120/10^2

Solve:

r = 1.2

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

6 0

3 years ago

The line tangent to the graph of g(x) = x^{3}-4x+1 at the point (2, 1) is given by the formula

Usimov [2.4K]

well, about A and D, I just plugged the values on the slope formula of

$\bf \begin{array}{llll} g(x)=x^3-4x+1\\ L(x) = 8(x-2)+1 \end{array} \qquad \begin{cases} x_1=1.9\\ x_2=2.1 \end{cases}\implies \cfrac{f(b)-f(a)}{b-a}$

for A the values are 8.01 and 8.0, so indeed those "slopes" are close. $\textit{\huge \checkmark}$

for D the values are -2.25 and 8.0, so no dice on that one.

for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.

checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice. $\textit{\huge \checkmark}$

for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.

for E, do they intercept at (2,1)? well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent. $\textit{\huge \checkmark}$

for F, we know the slope of the line L(x) is 8, is g'(2) = 8? let's check

recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice. $\textit{\huge \checkmark}$

6 0

3 years ago