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3 years ago

Carlos's ferry ride started at 5:55 p.m. The ferry ride ended at 7:01 p.m.

Mathematics

1 answer:

Serhud [2]3 years ago

8 0

Answer:1 hour 6 minutes

Step-by-step explanation:

5 min from 5:55 til 6:00. Then one whole hour from 6:00 to 7:00. Then, add on that extra 1 min. 1 hour and 6 min.

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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:

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

If a sweater sells for $ 48 after a 25 % markdown what was the original price

stiks02 [169]

The original price was $64
Hope this helps:)

7 0

3 years ago

Read 2 more answers

Which explanation correctly solves this problem?

Molodets [167]

Hello!

The Correct Answer for this Question would be:

"Start with 10.5. Add 3.5 and 10.5. Add 2.25 to the sum. Then add 5 to the answer."

In Other Words, Option "B".

Hope this Helps! Have A Wonderful Day! :)

8 0

4 years ago

Read 2 more answers

Maria’s phone bills were 95 67 43 and 115 what is the mean or average of her phone bill

Vesna [10]

Answer:

Step-by-step explanation:

You have to add all of the numbers together and divide them by how many numbers there are.

95 + 67 + 43 + 115 = 320

320 / 4 = 80

5 0

3 years ago