fourfold: Fourfold Table

Exercise template for computing joint probabilities from a 2x2 table based on three randomly-drawn conditional or marginal probabilities.

Name:
fourfold
Type:
Related:
Preview:

An industry-leading company seeks a qualified candidate for a management position. A management consultancy carries out an assessment center which concludes in making a positive or negative recommendation for each candidate: From previous assessments they know that of those candidates that are actually eligible for the position (event \(E\)) \(66\%\) get a positive recommendation (event \(R\)). However, out of those candidates that are not eligible \(65\%\) get a negative recommendation. Overall, they know that only \(9\%\) of all job applicants are actually eligible.

What is the corresponding fourfold table of the joint probabilities? (Specify all entries in percent.)

\(P(E \cap R)\)

\(P(\overline{E} \cap R)\)

\(P(E \cap \overline{R})\)

\(P(\overline{E} \cap \overline{R})\)

Using the information from the text, we can directly calculate the following joint probabilities: \[ \begin{aligned} P(E \cap R) & = P(R | E) \cdot P(E) = 0.66 \cdot 0.09 = 0.0594 = 5.94\%\\ P(\overline{E} \cap \overline{R}) & = P(\overline{R} | \overline{E}) \cdot P(\overline{E}) = 0.65 \cdot 0.91 = 0.5915 = 59.15\%. \end{aligned} \] The remaining probabilities can then be found by calculating sums and differences in the fourfold table:

\(R\) \(\overline{R}\) sum
\(E\) 5.94 3.06 9.00
\(\overline{E}\) 31.85 59.15 91.00
sum 37.79 62.21 100.00
  • \(P(E \cap R) = 5.94\%\)
  • \(P(\overline{E} \cap R) = 31.85\%\)
  • \(P(E \cap \overline{R}) = 3.06\%\)
  • \(P(\overline{E} \cap \overline{R}) = 59.15\%\)

An industry-leading company seeks a qualified candidate for a management position. A management consultancy carries out an assessment center which concludes in making a positive or negative recommendation for each candidate: From previous assessments they know that of those candidates that are actually eligible for the position (event \(E\)) \(63\%\) get a positive recommendation (event \(R\)). However, out of those candidates that are not eligible \(65\%\) get a negative recommendation. Overall, they know that only \(13\%\) of all job applicants are actually eligible.

What is the corresponding fourfold table of the joint probabilities? (Specify all entries in percent.)

\(P(E \cap R)\)

\(P(\overline{E} \cap R)\)

\(P(E \cap \overline{R})\)

\(P(\overline{E} \cap \overline{R})\)

Using the information from the text, we can directly calculate the following joint probabilities: \[ \begin{aligned} P(E \cap R) & = P(R | E) \cdot P(E) = 0.63 \cdot 0.13 = 0.0819 = 8.19\%\\ P(\overline{E} \cap \overline{R}) & = P(\overline{R} | \overline{E}) \cdot P(\overline{E}) = 0.65 \cdot 0.87 = 0.5655 = 56.55\%. \end{aligned} \] The remaining probabilities can then be found by calculating sums and differences in the fourfold table:

\(R\) \(\overline{R}\) sum
\(E\) 8.19 4.81 13.00
\(\overline{E}\) 30.45 56.55 87.00
sum 38.64 61.36 100.00
  • \(P(E \cap R) = 8.19\%\)
  • \(P(\overline{E} \cap R) = 30.45\%\)
  • \(P(E \cap \overline{R}) = 4.81\%\)
  • \(P(\overline{E} \cap \overline{R}) = 56.55\%\)

An industry-leading company seeks a qualified candidate for a management position. A management consultancy carries out an assessment center which concludes in making a positive or negative recommendation for each candidate: From previous assessments they know that of those candidates that are actually eligible for the position (event \(E\)) \(73\%\) get a positive recommendation (event \(R\)). However, out of those candidates that are not eligible \(62\%\) get a negative recommendation. Overall, they know that only \(12\%\) of all job applicants are actually eligible.

What is the corresponding fourfold table of the joint probabilities? (Specify all entries in percent.)

\(P(E \cap R)\)

\(P(\overline{E} \cap R)\)

\(P(E \cap \overline{R})\)

\(P(\overline{E} \cap \overline{R})\)

Using the information from the text, we can directly calculate the following joint probabilities: \[ \begin{aligned} P(E \cap R) & = P(R | E) \cdot P(E) = 0.73 \cdot 0.12 = 0.0876 = 8.76\%\\ P(\overline{E} \cap \overline{R}) & = P(\overline{R} | \overline{E}) \cdot P(\overline{E}) = 0.62 \cdot 0.88 = 0.5456 = 54.56\%. \end{aligned} \] The remaining probabilities can then be found by calculating sums and differences in the fourfold table:

\(R\) \(\overline{R}\) sum
\(E\) 8.76 3.24 12.00
\(\overline{E}\) 33.44 54.56 88.00
sum 42.20 57.80 100.00
  • \(P(E \cap R) = 8.76\%\)
  • \(P(\overline{E} \cap R) = 33.44\%\)
  • \(P(E \cap \overline{R}) = 3.24\%\)
  • \(P(\overline{E} \cap \overline{R}) = 54.56\%\)
Description:
Computing all four joint probabilities (in percent) in a fourfold table based on two randomly-generated conditional probabilities and one marginal probability. The exercise is a cloze with four numeric answers presented in a list.
Solution feedback:
Yes
Randomization:
Random numbers
Mathematical notation:
Yes
Verbatim R input/output:
No
Images:
No
Other supplements:
No
Raw: (1 random version)
PDF:
fourfold-Rmd-pdf
fourfold-Rnw-pdf
HTML:
fourfold-Rmd-html
fourfold-Rnw-html

(Note that the HTML output contains mathematical equations in MathML, rendered by MathJax using ‘mathjax = TRUE’. Instead it is also possible to use ‘converter = “pandoc-mathjax”’ so that LaTeX equations are rendered by MathJax directly.)

Demo code:

library("exams")

set.seed(403)
exams2html("fourfold.Rmd", mathjax = TRUE)
set.seed(403)
exams2pdf("fourfold.Rmd")

set.seed(403)
exams2html("fourfold.Rnw", mathjax = TRUE)
set.seed(403)
exams2pdf("fourfold.Rnw")