The following table presents the probability distribution of the number of vacations $X$ taken last year for a randomly chosen family. Find $P$ ( 1 or more).
$\begin{array}{c|ccccc}
x & 0 & 1 & 2 & 3 & 4 \
\hline P(x) & 0.05 & 0.69 & 0.16 & 0.08 & 0.02
\end{array}$

A) 0.69
B) 0.74
C) 0.95
D) 0.26

Question

The following table presents the probability distribution of the number of vacations $X$ taken last year for a randomly chosen family. Find $P$ ( 1 or more).
$\begin{array}{c|ccccc}
x & 0 & 1 & 2 & 3 & 4 \
\hline P(x) & 0.05 & 0.69 & 0.16 & 0.08 & 0.02
\end{array}$

A) 0.69
B) 0.74
C) 0.95
D) 0.26

Quizplus · Accepted Answer

To find $P$ (1 or more), we sum the probabilities of taking 1, 2, 3, or 4 vacations, which are 0.69, 0.16, 0.08, and 0.02, respectively. The sum is $0.69 + 0.16 + 0.08 + 0.02 = 0.95$.

The Following Table Presents the Probability Distribution of the Number XXX

The Following Table Presents the Probability Distribution of the Number $X$