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Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

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

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

I need answer for question 13

Mademuasel [1]

730/27 or improper fractions 27 1/27

6 0

3 years ago

Help ASAP plsssssssssssssssssssssss

Maurinko [17]

Answer:

Step-by-step explanation: idk

7 0

3 years ago

Read 2 more answers

Dose anyone know this problem if u so plz help!

Dennis_Churaev [7]

$A=\dfrac{1}{2}(a+b)h+\dfrac{1}{2}\pi r%2\\
a=8\\
b=4\\
h=4\\
r=2\\
A=\dfrac{1}{2}(8+4)\cdot4+\dfrac{1}{2}\pi \cdot 2^2\\
A=24+2\pi=2(12+\pi)$

7 0

4 years ago

Identify Relationships. Do the numbers show a pattern? 11.23,10.75,10.3,9.82,9.37,8.89

Alexxandr [17]

<h3>Answer:</h3>

Yes. The pattern is predicted by the formula ...

f(x) = -0.465x +11.68 +0.015·mod(x, 2)

<h3>Step-by-step explanation:</h3>

<em>Initial investigation</em>

The predictions from a line of best fit alternate being too low and too high. That is to say, every other point fits perfectly on a line, and the alternate points fit perfectly on a parallel line.

<em>Line through alternate points</em>

If we use x=1, 2, 3, ... for the number of the member of the sequence, then we have ....

(x, f(x)) = (2, 10.75), (4, 9.82), (6, 8.89)

These points describe a line that can be found using the two-point form of the equation for a line:

y = (y2 -y1)/(x2 -x1)·(x -x1) +y1

y = (9.82 -10.75)/(4 -2)·(x -2) +10.75 . . . . . using the values from the first two even-numbered points

y = (-0.93/2)·(x -2) +10.75 . . . . . . . . . . simplify a bit

y = -0.465x +11.68 . . . . . . . . . . . . . . . . simplify to slope-intercept form

<em>Correction for other points</em>

We know that this predicts a value too low when x is odd. We can find the amout the prediction is low and add a conditional term to make up the difference.

For x = 1,

y = -0.465·1 +11.68 = 11.215

The actual value is 11.23, or 11.23 -11.215 = 0.015 higher than the equation predicts. We can use the modulo function to give a value that is 1 for odd numbers and 0 for even numbers. For odd values of x, we can add this difference to make the formula predict every number in the sequence.

f(x) = -0.465x +11.68 +0.015·mod(x, 2) . . . . equation for the given points

_____

The table attached shows the given points (y1) and the values predicted using the above formula (f(x1)). (The x1 and y1 used in the attachment are the names of the lists of x-values and y-values. They are not related to the x1 and y1 used above in the 2-point form of the equation for a line.)

6 0

3 years ago