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).
$\begin{array} { c | c c c c c }
x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & 0.09 & 0.63 & 0.18 & 0.07 & 0.03
\end{array}$
A) 0.63
B) 0.72
C) 0.28
D) 0.91

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). 
$\begin{array} { c | c c c c c } 
x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & 0.09 & 0.63 & 0.18 & 0.07 & 0.03
\end{array}$
A) 0.63
B) 0.72
C) 0.28
D) 0.91

Quizplus · Accepted Answer

To find $P(1 \text{ or more})$, we sum the probabilities of taking 1, 2, 3, or 4 vacations: $0.63 + 0.18 + 0.07 + 0.03 = 0.91$.

The Following Table Presents the Probability Distribution of the Number x01234P(x)0.090.630.180.070.03\begin{array} { c | c c c c c } x & 0 & 1 & 2 & 3 & 4 \\\hline P ( x ) & 0.09 & 0.63 & 0.18 & 0.07 & 0.03\end{array}xP(x)​00.09​10.63​20.18​30.07​40.03​​

The Following Table Presents the Probability Distribution of the Number $\begin{array} { c | c c c c c } x & 0 & 1 & 2 & 3 & 4 \\\hline P ( x ) & 0.09 & 0.63 & 0.18 & 0.07 & 0.03\end{array}$