It is very common for television series to draw a large audience for special events of for cliff-hanging story lines. suppose that on one of these occasions, the special show drew viewers from 38.2% of all us tv households. suppose that three tv households are randomly selected. what is the probability that all three households viewed this special show?
The given case satisfies the condition of Binomial Experiment so we can use the Binomial Probability to answer this question. Conditions for a Binomial Experiment are: 1) The sample is simple random sample 2) The probability of success is constant 3) The trials are independent 4) There are a fixed number of trials
Probability of success on a single trial = p = 0.382 Number of trials = n = 3 Number of successful trials= x = 3
The formula of the Binomial probability is: P( 3 out of 3 success) = [tex]3C_{3} p^{3} (1-p)^{3-3} [/tex] Using the values, we get:
P (3 out of 3) = 0.557 = 5.57% Thus the probability that all three households viewed this special show is 5.57%
