Hi! Short math question

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

2 years ago

In a certain town, 22% of voters favor a given ballot measure. for groups of 21 voters, find the variance for the number who fav

Galina-37 [17]

c. 4.6

21 X .22= 4.6

Calculating the variance requires finding the product of 21 and 22%. To make this easier we convert 22% into it's decimal form and construct the equation. To back check this answer we can use 10% of 21 voters which equals 2.1% then double that amount to reach 4.2%, knowing that we now have a close approximation of the variance we can eliminate answers a, b, and d, leaving c as the only logical choice.

5 0

2 years ago

A rectangular field is 30 yards in length and 81 feet in width.

PtichkaEL [24]

Answer:

7290in ^2

Step-by-step explanation:

covert 30yd to feet which equals too 90ft.

Then multiple 90 x 81 = 7290

8 0

1 year ago

Simplify the given expression. (p+6)(p-4)

vampirchik [111]

Answer:p2+2p−24

Step-by-step explanation:(p+6)(p−4)

=(p+6)(p+−4)

=(p)(p)+(p)(−4)+(6)(p)+(6)(−4)

=p2−4p+6p−24

=p2+2p−24

8 0

3 years ago

Read 2 more answers

Maya made $324 for 18 hours of work. At the same rate how many hours would she have to work to make $126

Illusion [34]

Answer:7

Step-by-step explanation:

324/18 hours = 18$ per hour

126/18$=7 hours

4 0

3 years ago

Read 2 more answers