Compare 70.000+1,000+20+3 to 94119

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:

B = bugs are present in a computer program.

D = 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 E given that another event X 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

Find the vertex and length of the latus rectum for the parabola. y=1/6(x-8)^2+6

Ivan

Step-by-step explanation:

If the parabola has the form

$y = a(x - h)^2 + k$ (vertex form)

then its vertex is located at the point (h, k). Therefore, the vertex of the parabola

$y = \dfrac{1}{6}(x - 8)^2 + 6$

is located at the point (8, 6).

To find the length of the parabola's latus rectum, we need to find its focal length f. Luckily, since our equation is in vertex form, we can easily find from the focus (or focal point) coordinate, which is

$\text{focus} = (h, k +\frac{1}{4a})$

where $\frac{1}{4a}$ is called the focal length or distance of the focus from the vertex. So from our equation, we can see that the focal length f is

$f = \dfrac{1}{4(\frac{1}{6})} = \dfrac{3}{2}$

By definition, the length of the latus rectum is four times the focal length so therefore, its value is

$\text{latus rectum} = 4\left(\dfrac{3}{2}\right) = 6$

5 0

3 years ago

A project is graded on a scale of 1 to 5. If the random variable, X, is the project grade, what is the mean of the probability

alex41 [277]

Answer:

the mean is 3

answer choice D

Step-by-step explanation:

60/20

8 0

3 years ago

Read 2 more answers

Which is longer 2 meters or 200 cent a meters or 2.05 meters

Yuri [45]

Solving this requires use of conversions

we already know that
2.05 m > 2 m

all that's left is the question
200 cm vs. 2.05 meters

I we know our equivalencies, we know that
100 cm = 1 m

if this is so then
200 cm = 2 m

we already know that
2.05 m > 2 m

so it must also be that
2.05 m > 200 cm

$2.05 \: meters$

6 0

3 years ago

Does anybody have the answers to this

Vinvika [58]

I cant read it srry is wrinkled

3 0

3 years ago