Given the contingency table shown here, find P(A₃?B₂). $$\begin{array} { |c | c | c | c | c | c | } \hline{ } & { A _ { 1 } } & A _ { 2 } & A _ { 3 } & A _ { 4 } & \text { Row Total } \ \hline B _ { 1 } & 12 & 26 & 42 & 68&148 \ \hline B _ { 2 } & 14 & 28 & 44 & 64 &150\ \hline B _ { 3 } & 18 & 32 & 47 & 72 &169\ \hline \text { Col Total } & 44 & 86 & 133 & 204 & 467 \ \hline \end{array}$$ A).3212 B).2933 C).0942 D).1006

Question

Quizplus · Accepted Answer

To find P(A₃∩B₂), we first need to determine the values in the cells that correspond to the events A₃ and B₂. From the contingency table, we see that the total number of observations is 467. The event A₃ corresponds to the cell with row A1 and column B3, which has a count of 42. The event B₂ corresponds to the cell with row B1 and column B2, which has a count of 26. Therefore, P(A₃∩B₂) = 42/467 × 26/467 = 1092/218089 ≈ 0.005 = 0.94%. The closest choice to this value is C) 0.0942, so the correct answer is C).

Given the Contingency Table Shown Here, Find P(A3?B2) A)3212 B)

Given the Contingency Table Shown Here, Find P(A₃?B₂) A)3212
B)