Answer:
(a) The value of P (None) is 0.062.
(b) The value of P(at least one) is 0.938.
(c) The value of P(at most one) is 0.253.
(d) The event is not unusual.
Step-by-step explanation:
Let <em>X</em> = number of households watching the show.
The probability of the random variable <em>x</em> is, P (X) = <em>p</em> = 0.18.
The sample selected for the survey is of size, <em>n</em> = 14
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 14 and <em>p</em> = 0.18.
The probability of a Binomial distribution is computed using the formula:

(a)
Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

Thus, the value of P (None) is 0.062.
(b)
Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)

Thus, the value of P(at least one) is 0.938.
(c)
Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)

Thus, the value of P(at most one) is 0.253.
(d)
An event that has a very low probability of occurrence is known as an unusual event.
The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.
This probability value is not low.
Hence, the event is not unusual.