Use the probability distribution to complete parts (a) through (d) below.
The probability distribution of number of televisions per household in a small town
$\begin{array}{lccccc}\mathbf{x} & 0 & 1 & 2 & 3 & \text { 믄 } \\ \mathrm{P}(\mathbf{x}) & 0.02 & 0.16 & 0.25 & 0.57\end{array}$
The probability is
(Type an integer or a decimal. Do not round.)
(b) Find the probability of randomly selecting a household that has two or more televisions.
The probability is
(Type an integer or a decimal. Do not round.)
(c) Find the probability of randomly selecting a household that has between one and three televisions, inclusive.
The probability is
(Type an integer or a decimal. Do not round.)
(d) Find the probability of randomly selecting a household that has at most two televisions.
Answer
Final Answer: (a) The probability of randomly selecting a household that has two or more televisions is \(\boxed{0.82}\). (b) The probability of randomly selecting a household that has between one and three televisions, inclusive, is \(\boxed{0.98}\). (c) The probability of randomly selecting a household that has at most two televisions is \(\boxed{0.43}\).
Step 1 ：The probability distribution of the number of televisions per household in a small town is given as follows: \(P(0) = 0.02\), \(P(1) = 0.16\), \(P(2) = 0.25\), and \(P(3) = 0.57\).
Step 2 ：For part (a), we need to find the probability of randomly selecting a household that has two or more televisions. This can be found by adding the probabilities for 2 and 3 televisions: \(P(2) + P(3) = 0.25 + 0.57 = 0.82\).
Step 3 ：For part (b), we need to find the probability of randomly selecting a household that has between one and three televisions, inclusive. This can be found by adding the probabilities for 1, 2, and 3 televisions: \(P(1) + P(2) + P(3) = 0.16 + 0.25 + 0.57 = 0.98\).
Step 4 ：For part (c), we need to find the probability of randomly selecting a household that has at most two televisions. This can be found by adding the probabilities for 0, 1, and 2 televisions: \(P(0) + P(1) + P(2) = 0.02 + 0.16 + 0.25 = 0.43\).
Step 5 ：Final Answer: (a) The probability of randomly selecting a household that has two or more televisions is \(\boxed{0.82}\). (b) The probability of randomly selecting a household that has between one and three televisions, inclusive, is \(\boxed{0.98}\). (c) The probability of randomly selecting a household that has at most two televisions is \(\boxed{0.43}\).
