Jasmine was 1.5 m tall. In the last couple of years, her height has increased by 20%. How tall is Jasmine now?

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

Find the value of two numbers is there some is 12 and there difference is 4

Mademuasel [1]

You mean their sum is 12?

A) x + y = 12
B) x -y = 4 then we add both equations:
2x = 16
x = 8
y = 4

3 0

3 years ago

Seven years ago, Douglas purchased a $204,000 home with a 30-year

Stella [2.4K]

Answer:

The new mortgage repayment is $10,580.69

Step-by-step explanation:

As at the time Douglas lost his job,he has already made repayments on the mortgage for a period of 7 years,by extending the repayment period by another 7 years,Douglas now have a period of repayment of 30 years now(30-7+7).

The new repayment yearly can be computed using the pmt formula in excel as given below:

=pmt(rate,nper,-pv,fv)

rate is the interest rate on mortgage given as 4.25%

nper is the period of repayment now 30 years

pv is the current of balance of $177,533.62

fv is the total amount repayable on the mortgage and it is not known hence taken as zero

=pmt(4.25%,30,-177533.62,0)=$10,580.69

8 0

3 years ago

Read 2 more answers

I don't know this can someone help me?

Gemiola [76]