Suppose that you move to a new house and you are 10% sure that your new house's phone number is 561290. To verify this, you use the house's phone to dial 561290, obtain a busy signal, and conclude that this is indeed your phone number. (Suppose that you were incorrect, and that this is somebody else's phone number. In that case, you will receive a busy signal if and only if that person is making a phone call at the same time. But if you call your own number, you will always receive a busy signal.)

Question

Suppose that you move to a new house and you are 10% sure that your new house's phone number is 561290. To verify this, you use the house's phone to dial 561290, obtain a busy signal, and conclude that this is indeed your phone number. (Suppose that you were incorrect, and that this is somebody else's phone number. In that case, you will receive a busy signal if and only if that person is making a phone call at the same time. But if you call your own number, you will always receive a busy signal.)

Assume that the probability of a random six-digit phone number being in use at any given time is 0.001. What is the probability that you will be correct in concluding that 561290 is indeed your own phone number? Your answer should be accurate to 3 decimal places.

Answer 1

To calculate the probability that you will be correct in concluding that 561290 is your own phone number, we can use Bayes' theorem.

Let's define the events:
A: The house's phone number is 561290.
B: You dial 561290 and receive a busy signal.

We need to find the probability of event A given event B, P(A|B).

According to the problem, you are 10% sure that the phone number is 561290, so we have P(A) = 0.10.

The probability of receiving a busy signal when the phone number is not yours is 0.001 (given in the question), so P(B|not A) = 0.001.

The probability of receiving a busy signal when the phone number is yours is 1 (given in the question), so P(B|A) = 1.

We can calculate P(not A) using the complement rule since P(A) + P(not A) = 1:
P(not A) = 1 - P(A) = 1 - 0.10 = 0.90.

Now we can use Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / (P(B|A) * P(A) + P(B|not A) * P(not A))
= (1 * 0.10) / (1 * 0.10 + 0.001 * 0.90)
= 0.10 / (0.10 + 0.0009)
= 0.10 / 0.1009
≈ 0.992 (rounded to 3 decimal places).

Therefore, the probability that you will be correct in concluding that 561290 is indeed your own phone number is approximately 0.992.

Answer 2

To find the probability that you will be correct in concluding that 561290 is your own phone number, we need to calculate the conditional probability.

Let's define the probabilities:

P(A) = Probability that your new house's phone number is actually 561290 (given as 10%, so P(A) = 0.1)
P(B) = Probability that a random six-digit phone number is in use at any given time (given as 0.001, so P(B) = 0.001)
P(B') = Probability that a random six-digit phone number is NOT in use at any given time (obtained by subtracting P(B) from 1, so P(B') = 1 - P(B) = 1 - 0.001 = 0.999)

We are looking for P(A|B), which is the probability that your new house's phone number is 561290 given that you receive a busy signal when you call that number.

Using Bayes' Theorem, we have:

P(A|B) = (P(B|A) * P(A)) / P(B)

P(B|A) is the probability that you receive a busy signal when you call your own number (which is always true), so P(B|A) = 1.
P(A|B) = (1 * 0.1) / P(B)

To calculate P(A|B), we need to find P(B), the probability that you receive a busy signal when you call any number.

Since P(B) is the probability that a random six-digit phone number is in use at any given time, it is given as 0.001.

Now we can solve for P(A|B):

P(A|B) = (1 * 0.1) / 0.001
P(A|B) = 0.1 / 0.001
P(A|B) = 100

Therefore, the probability that you will be correct in concluding that 561290 is your own phone number is 100 or 100%.