Question

The probability that a contractor will get a plumbing contract is \(2/3\) and the probability that he will not get an electric contract is \(5/9\). If the probability of getting at least one contract is \(4/5\), what is the probability that he will get both?

• A. \(31/45\)
• B. \(8/45\)
• C. \(14/45\)
• D. None of these

Hint:

"At least one" translates to a union; immediately think of \(P(A \cup B)=P(A)+P(B)-P(A\cap B)\). Alternatively, complement method often works for "at least"/"at most" wordings.

Answer 1

The Correct Option is C

Step 1: Define events and given probabilities.
Let \(P\) = gets Plumbing, \(E\) = gets Electric.
\(P(P) = \dfrac{2}{3}\).
Given \(P(\overline{E}) = \dfrac{5}{9} \Rightarrow P(E) = 1 - \dfrac{5}{9} = \dfrac{4}{9}\).
"At least one" means \(P(P \cup E) = \dfrac{4}{5}\). Step 2: Use the union formula to find the intersection.
\(P(P \cup E) = P(P) + P(E) - P(P \cap E)\).
So \(P(P \cap E) = P(P) + P(E) - P(P \cup E)\).
\(\Rightarrow P(P \cap E) = \dfrac{2}{3} + \dfrac{4}{9} - \dfrac{4}{5}\).
With denominator \(45\): \(\dfrac{2}{3} = \dfrac{30}{45},\ \dfrac{4}{9} = \dfrac{20}{45},\ \dfrac{4}{5} = \dfrac{36}{45}\).
\(\Rightarrow P(P \cap E) = \dfrac{30+20-36}{45} = \dfrac{14}{45}\). Step 3: Conclude.
Probability of getting both \(= \boxed{\dfrac{14}{45}}\).