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Zielflug [23.3K]
3 years ago
12

A tortilla chip has a mass of about 2 grams. Find the mass in milligrams.

Mathematics
1 answer:
Pani-rosa [81]3 years ago
7 0
I love me some tortilla chips :)
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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
2/3 - 1/5 as a fraction
Goryan [66]

Answer:

Step-by-step explanation:

8 0
3 years ago
Please help need this for today
Cloud [144]
All will apply except for (A, C, and E) because those lengths are less than 4/6. So (B, D and F) are the correct answers.
6 0
3 years ago
What is 5d - 10 = 20​
algol [13]

Answer:

d=6

Step-by-step explanation:

move all terms that don't contain d to the right side and solve.

6 0
3 years ago
Read 2 more answers
Find the value of x<br> 2x + 46 = 7x + 6
Vlad1618 [11]

Answer:

x = 8

Step-by-step explanation:

2x + 46 = 7x + 6

Add -2x and -6 on both sides.

46 - 6 = 7x - 2x

40 = 5x

Divide 5 into both sides.

40/5 = x

8 = x

8 0
3 years ago
Read 2 more answers
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