Which correctly uses bar notation to represent the repeating decimal for 6/11?

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

Please help ASAP! Due in a hour! :) answer both and I will thank and mark as brainliest

lorasvet [3.4K]

Answer:

22: 45 and 126

Step-by-step explanation:

6 0

2 years ago

How many times dose 18 go into 170

Leokris [45]

I think it is 10 because 18×10=180 but it is close hope this helps

7 0

3 years ago

What is the slope of the line that passes through the points (−10,−8) and (-8, -16) Write your answer in simplest form.

jarptica [38.1K]

Step-by-step explanation:

given,

point = ( -10, -8 ) and ( -8, -16 )

we know,

slope = y2 - y1 / x2 - x1

x1 , y1 = (-10 , -8 )

x2 , y2 = ( -8, -16 )

after inserting the values we got,

→ -16 - (-8 ) / -8 - (-10)

→ -8/2 → -4

therefore, slope of line is -4.

hope this answer helps you dear...take care and may this answer helps you dear!.

7 0

2 years ago

Help me with this please everyone

cupoosta [38]

First, let's simplify each expression by combining like terms:

A: 9x-3x-4 = 6x-4

B: 12x-4 = 12x-4 (Already Combined terms)

C: 5x+x-4 = 6x-4

When looking at the list of equations, it is clear that only Expression A and C are equivalent. Brianna failed to combine like terms and plug in more points than 0.

I Hope this Helps!
-Sinnamin

3 0

3 years ago