10 gallons equals how many liters

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

List the next four multiples of the fraction. 3/6

Arlecino [84]

They are 6/6, 9/6, 12/6, and 15/6. Please give brainliest!!!

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2 years ago

Read 2 more answers

A rectangular plot measures 20 ft. By 30ft. A 3ft wide side walk surrounds it. Find the area of the side walk

r-ruslan [8.4K]

Answer:

336 feet²

Step-by-step explanation:

If we have a rectangle that is 30 by 20 feet, that means the area of that rectangle would be 20 × 30 feet squared, which is 600 ft².

If there is a 3 feet sidewalk surrounding it, that means that the end of the sidewalk will extend 3 feet extra around each side of plot. Since there are two ends to one side, that means an extra six feet is added on to each dimension. Therefore, 36 × 26 are the dimensions of the sidewalk+plot. 36 × 26 = 936 ft².

To find the area of the sidewalk itself, we subtract 600 ft² from 936 ft². This gets us with 336 ft².

Hope this helped!

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What is the volume of this cone?

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Answer:

The answer is A 41.9 feet.

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(5.27) An owner of a home in the Midwest installed solar panels to reduce heating costs. After installing the solar panels, he m

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The regression line equation is $y=85+16x$ , the value of the slope is shown as constant 16 and it's a positive constant, hence the correlation between the amount of gas used and degree of the days is a POSITIVE correlation.

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