Answer:
(a) The probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.
(b) The probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.
Step-by-step explanation:
Let's denote the events as follows:
A = Fell short of expectations
B = Met expectations
C = Surpassed expectations
N = no response
Given:
P (N) = 0.04
P (A) = 0.26
P (B) = 0.65
(a)
Compute the probability that a randomly selected alumnus would say their experience surpassed expectations as follows:
[tex]P(C) = 1 - [P(A) + P(B) + P(N)]\\= 1 - [0.26 + 0.65 + 0.04]\\= 1 - 0.95\\= 0.05[/tex]
Thus, the probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.
(b)
The response of all individuals are independent.
Compute the probability that a randomly selected alumnus would say their experience met or surpassed expectations as follows:
[tex]P(B\cup C) = P(B)+P(C)-P(B\cap C)\\=P(B)+P(C)-P(B)\times P(C)\\= 0.65 + 0.05 - (0.65\times0.05)\\=0.6675\\\approx0.67[/tex]
Thus, the probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.