The following table presents the probability distribution of the number of vacations X taken last year for a randomly chosen family. Find the probability that a family took at least 3 vacations last year.

$\begin{array} { c | c c c c c }
x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & 0.08 & 0.66 & 0.15 & 0.09 & 0.02
\end{array}$

A) 0.11
B) 0.09
C) 0.26
D) 0.89

Question

The following table presents the probability distribution of the number of vacations X taken last year for a randomly chosen family. Find the probability that a family took at least 3 vacations last year.

\begin{array} { c | c c c c c } x & 0 & 1 & 2 & 3 & 4 \\\hline P ( x ) & 0.08 & 0.66 & 0.15 & 0.09 & 0.02\end{array}

$\begin{array} { c | c c c c c } 
x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & 0.08 & 0.66 & 0.15 & 0.09 & 0.02
\end{array}$

A) 0.11
B) 0.09
C) 0.26
D) 0.89

Quizplus · Accepted Answer

The probability that a family took at least 3 vacations last year is the sum of the probabilities for taking 3 and 4 vacations, which is $0.09 + 0.02 = 0.11$.

The Following Table Presents the Probability Distribution of the Number x01234P(x)0.080.660.150.090.02\begin{array} { c | c c c c c } x & 0 & 1 & 2 & 3 & 4 \\\hline P ( x ) & 0.08 & 0.66 & 0.15 & 0.09 & 0.02\end{array}xP(x)​00.08​10.66​20.15​30.09​40.02​​

The Following Table Presents the Probability Distribution of the Number $\begin{array} { c | c c c c c } x & 0 & 1 & 2 & 3 & 4 \\\hline P ( x ) & 0.08 & 0.66 & 0.15 & 0.09 & 0.02\end{array}$