Question

You are a meteorologist that places temperature sensors all of the world, and you set them up so that they automatically e-mail you, each day, the high temperature for that day. Unfortunately, you have forgotten whether you placed a certain sensor S in Maine or in the Sahara desert (but you are sure you placed it in one of those two places) . The probability that you placed sensor S in Maine is 5%. The probability of getting a daily high temperature of 80 degrees or more is 20% in Maine and 90% in Sahara. Assume that probability of a daily high for any day is conditionally independent of the daily high for the previous day, given the location of the sensor.

Part a : If the first e-mail you got from sensor S indicates a daily high under 80 degrees, what is the probability that the sensor is placed in Maine?
Part b : If the first e-mail you got from sensor S indicates a daily high under 80 degrees, what is the probability that the second e-mail also indicates a daily high under 80 degrees?
Part c: What is the probability that the first three e-mails all indicate daily highs under 80 degrees?

Answer 1

Part a: The probability that the sensor is placed in Maine is approximately 34.78%. Part b: The probability that the second email also indicates a daily high under 80 degrees is approximately 13.5%. Part c: The probability that the first three emails all indicate daily highs under 80 degrees is approximately 0.2437%.

Step-by-step explanation:

Part a:

To find the probability that the sensor is placed in Maine given that the first email indicates a daily high under 80 degrees, we can use Bayes' theorem.

Let P(M) be the probability that the sensor is placed in Maine, and P(S) be the probability that the sensor reports a daily high under 80 degrees.

Using Bayes' theorem, we have:

P(M|S) = (P(S|M) * P(M)) / P(S)

We are given P(M) = 0.05, P(S|M) = 0.8, and P(S) = P(S|Maine) * P(Maine) + P(S|Sahara) * P(Sahara) = (0.2 * 0.05) + (0.1 * 0.95) = 0.02 + 0.095 = 0.115.

Substituting the values, we get:

P(M|S) = (0.8 * 0.05) / 0.115 = 0.04 / 0.115 = 0.3478.

Therefore, the probability that the sensor is placed in Maine given that the first email indicates a daily high under 80 degrees is approximately 0.3478 or 34.78%.

Part b:

To find the probability that the second email also indicates a daily high under 80 degrees, we need to consider the conditional probabilities.

If the sensor is placed in Maine, the probability of getting a daily high under 80 degrees is 0.8.

If the sensor is placed in the Sahara, the probability of getting a daily high under 80 degrees is 0.1.

Let P(D2) be the probability that the second email indicates a daily high under 80 degrees.

If the sensor is in Maine, P(D2|M) = 0.8.

If the sensor is in the Sahara, P(D2|S) = 0.1.

Using the law of total probability, we have:

P(D2) = P(D2|M) * P(M) + P(D2|S) * P(S) = 0.8 * 0.05 + 0.1 * 0.95 = 0.04 + 0.095 = 0.135.

Therefore, the probability that the second email also indicates a daily high under 80 degrees is 0.135 or 13.5%.

Part c:

To find the probability that the first three emails all indicate daily highs under 80 degrees, we can use the conditional independence of the daily high temperatures given the location of the sensor.

Let P(D1) be the probability that the first email indicates a daily high under 80 degrees.

If the sensor is in Maine, P(D1|M) = 0.8. If the sensor is in the Sahara, P(D1|S) = 0.1.

Using the law of total probability, we have:

P(D1) = P(D1|M) * P(M) + P(D1|S) * P(S) = 0.8 * 0.05 + 0.1 * 0.95 = 0.04 + 0.095 = 0.135.

Similarly, let P(D2) and P(D3) be the probabilities that the second and third emails indicate daily highs under 80 degrees, respectively.

Since the daily high temperatures are conditionally independent given the location of the sensor, we have:

P(D2) = P(D1)

P(D3) = P(D1)

Therefore, the probability that the first three emails all indicate daily highs under 80 degrees is (0.135)^3 = 0.00243675 or 0.2437%.