Given the contingency table shown here, find P(A or B). $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Age }$
$\begin{array} {| l | c | c | c| }

\hline\text { Absences } & \text { Under } 25 ( A ) & 25 \text { or More } \left( A ^ { \prime } \right) & \text { Row Total } \
\hline \text { Under } 2 \text { days } ( B ) & 50 & 40 & 90 \
\hline 2 \text { or more days } \left( B ^ { \prime } \right) & 30 & 80 & 110 \
\hline \text { Column Total } & 80 & 120 & 200\
\hline
\end{array}$
A).25
B).85
C).60
D).42

Question

Given the contingency table shown here, find P(A or B). $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Age }$
$\begin{array} {|  l | c | c | c|  }

\hline\text { Absences } & \text { Under } 25 ( A ) & 25 \text { or More } \left( A ^ { \prime } \right) & \text { Row Total } \
\hline  \text { Under } 2 \text { days } ( B ) & 50 & 40 & 90 \
\hline  2 \text { or more days } \left( B ^ { \prime } \right) & 30 & 80 & 110 \
\hline \text { Column Total } & 80 & 120 & 200\
\hline
\end{array}$
A).25
B).85
C).60
D).42

Quizplus · Accepted Answer

P(A or B) can be found by adding the probabilities of A and B and subtracting the probability of their intersection:

P(A or B) = P(A) + P(B) - P(A and B)

To find P(A), we add the probabilities of the two cells in row A:

P(A) = 50/200 + 80/200 = 0.65

To find P(B), we add the probabilities of the two cells in column B:

P(B) = 50/200 + 30/200 = 0.4

To find P(A and B), we need to look at the cell where row A and column B intersect:

P(A and B) = 50/200

Now we can substitute these values into the formula:

P(A or B) = 0.65 + 0.4 - 0.25 = 0.6

Therefore, the best choice is C, 0.6.

Given the Contingency Table Shown Here, Find P(A or B) \quad

Given the Contingency Table Shown Here, Find P(A or B) $\quad$