\(P\)(Adult needs glasses) = \(P\)(Adult needs glasses \(\cap\) Adult uses glasses) + \(P\)(Adult needs glasses \(\cap\) Adult doesn’t use glasses)
\(P\)(Adult uses glasses) = 0.44 + 0.14
\(P\)(Adult uses glasses) = 0.58
\(P\)(Adult needs glasses \(\cap\) Adult doesn’t use glasses) = 0.14
\(P\)(Adult uses glasses) = \(P\)(Adult uses glasses \(\cap\) Adult needs glasses) + \(P\)(Adult uses glasses \(\cap\) Adult doesn’t need glasses)
\(P\)(Adult uses glasses) = 0.44 + 0.02
\(P\)(Adult uses glasses) = 0.46
\(S\) = {income \(>\) 43,318, income \(\leq\) 43,318}
\(P\)(At least two had incomes exceeding 43,318) = \(P\)(Two had incomes exceeding 43,318) + \(P\)(Three had incomes exceeding 43,318) + \(P\)(Four had incomes exceeding 43,318)
\(P\)(Exactly two had incomes exceeding 43,318) is already a simple event
\(P\)(Exactly one had an income less than or equal to 43,318) is already a simple event
\(P\)(At least two had incomes exceeding 43,318) = \(P\)(Two had incomes exceeding 43,318) + \(P\)(Three had incomes exceeding 43,318) + \(P\)(Four had incomes exceeding 43,318)
\(P\)(At least two had incomes exceeding 43,318) = \({4 \choose 2}(0.5)^2(0.5)^2 + {4 \choose 3}(0.5)^3(0.5)^1 + {4 \choose 4}(0.5)^4(0.5)^0\)
\(P\)(At least two had incomes exceeding 43,318) = \(0.375 + 0.25 + 0.0625\)
\(P\)(At least two had incomes exceeding 43,318) = \(0.6875\)
\(P\)(Exactly two had incomes exceeding 43,318) = \({4 \choose 2}(0.5)^2(0.5)^2\)
\(P\)(Exactly two had incomes exceeding 43,318) = \(0.375\)
\(P\)(Exactly one had an income less than or equal to 43,318) = \({4 \choose 1}(0.5)^1(0.5)^3\)
\(P\)(Exactly one had an income less than or equal to 43,318) = \(0.25\)
\(A\) = Randomly selected employee scores a passing grade \(M\) = Randomly selected employee is male
Events \(A\) and \(M\) are independent if \(P\)(\(A|M\)) = \(P\)(\(A\)).
\(P\)(\(A|M\)) = \(\frac{24}{40}\)
\(P\)(\(A\)) = \(\frac{60}{100}\)
Both probabilities equal \(\frac{3}{5}\) or 0.6, so events \(A\) and \(M\) are independent.
\(\bar{A}\) = Randomly selected employee scores a failing grade \(F\) = randomly selected employee is female
Events \(\bar{A}\) and \(F\) are independent if \(P\)(\(\bar{A}|F\)) = \(P\)(\(\bar{A}\)).
\(P\)(\(\bar{A}|F\)) = \(\frac{24}{60}\)
\(P\)(\(\bar{A}\)) = \(\frac{40}{100}\)
Both probabilities equal \(\frac{2}{5}\) or 0.4, so events \(\bar{A}\) and \(F\) are independent.
It is impossible for \(P\)(\(A \cap B\)) = 0.1 as that would mean that \(P\)(\(A \cup B\)) = \(P\)(\(A\)) + \(P\)(\(B\)) - \(P\)(\(A \cap B\)) = \(0.7+0.8-0.1\) = \(1.4 > 1\), which would result in an event with a probability that is greater than one.
The greatest value of \(P\)(\(A \cup B\)) is 1, so the smallest value of \(P\)(\(A \cap B\)) in \(P\)(\(A\)) + \(P\)(\(B\)) - \(P\)(\(A \cap B\)) must be \(0.5\).
It is impossible for \(P\)(\(A \cap B\)) = 0.77 as that would mean that \(P\)(\(A \cap B\)) \(> P\)(\(A\)).
\(P\)(\(A \cap B\)) must be smaller than both \(P\)(\(A\)) and \(P\)(\(B\)), so the greatest value of \(P\)(\(A \cap B\)) must be \(0.7\).
\(P\)(Disease I \(\cup\) Disease II) = \(P\)(Disease I) + \(P\)(Disease II) - \(P\)(Disease I \(\cap\) Disease II)
\(P\)(Disease I \(\cup\) Disease II) = 0.10 + 0.15 - 0.03
\(P\)(Disease I \(\cup\) Disease II) = 0.22
\(P\)(Disease I \(\cap\) Disease II) | \(P\)(Disease I \(\cup\) Disease II) = \(P\)(Disease I \(\cap\) Disease II)/\(P\)(Disease I \(\cup\) Disease II)
\(P\)(Disease I \(\cap\) Disease II) | \(P\)(Disease I \(\cup\) Disease II) = 0.03/0.22 = \(\frac{3}{22}\)
\(P\)(Both positive) = \(0.95 \cdot 0.1 = 0.095\)
\(P\)(Lying positive and Truthful negative) = \(0.95 \cdot 0.9 = 0.855\)
\(P\)(Lying negative and Truthful positive) = \(0.05 \cdot 0.1 = 0.005\)
\(P\)(At least one positive) = \(1- P\)(Both negative) = \(1-(0.05 \cdot 0.9)\) = \(1-(0.045)\) = \(0.955\)