Understanding the Probability Problem

This question asks us to calculate the probability that a contractor receives both a plumbing contract and an electric contract, given the probabilities of getting each type of contract individually and the probability of getting at least one of them.

Let's define the events:

• Let event A be that the contractor gets the plumbing contract.
• Let event B be that the contractor gets the electric contract.

We are given the following probabilities:

• The probability of getting a plumbing contract, \(P(A)\), is \( \frac{2}{3} \).
• The probability of not getting an electric contract, \(P(B')\), is \( \frac{5}{9} \).
• The probability of getting at least one contract (either plumbing or electric or both), \(P(A \cup B)\), is \( \frac{4}{5} \).

We need to find the probability that the contractor will get both contracts, which is \(P(A \cap B)\).

Calculating the Probability of Getting the Electric Contract

We are given the probability of *not* getting the electric contract, \(P(B')\). The probability of getting the electric contract, \(P(B)\), is simply 1 minus the probability of not getting it.

Using the complementary probability rule:

\( P(B) = 1 - P(B') \)

Substitute the given value:

\( P(B) = 1 - \frac{5}{9} \)

\( P(B) = \frac{9}{9} - \frac{5}{9} \)

\( P(B) = \frac{9 - 5}{9} \)

\( P(B) = \frac{4}{9} \)

So, the probability of getting the electric contract is \( \frac{4}{9} \).

Using the Probability Formula for 'At Least One Event'

The probability of getting at least one contract (plumbing or electric) is given by the formula for the union of two events:

\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

We know \(P(A)\), \(P(B)\), and \(P(A \cup B)\). We can rearrange this formula to solve for \(P(A \cap B)\), the probability of getting both contracts:

\( P(A \cap B) = P(A) + P(B) - P(A \cup B) \)

Solving for the Probability of Getting Both Contracts

Now, substitute the known values into the rearranged formula:

\( P(A \cap B) = \frac{2}{3} + \frac{4}{9} - \frac{4}{5} \)

To add and subtract these fractions, we need to find a common denominator. The denominators are 3, 9, and 5. The least common multiple (LCM) of 3, 9, and 5 is 45.

Convert each fraction to an equivalent fraction with a denominator of 45:

• \( \frac{2}{3} = \frac{2 \times 15}{3 \times 15} = \frac{30}{45} \)
• \( \frac{4}{9} = \frac{4 \times 5}{9 \times 5} = \frac{20}{45} \)
• \( \frac{4}{5} = \frac{4 \times 9}{5 \times 9} = \frac{36}{45} \)

Now substitute these equivalent fractions back into the equation:

\( P(A \cap B) = \frac{30}{45} + \frac{20}{45} - \frac{36}{45} \)

Perform the addition and subtraction:

\( P(A \cap B) = \frac{30 + 20 - 36}{45} \)

\( P(A \cap B) = \frac{50 - 36}{45} \)

\( P(A \cap B) = \frac{14}{45} \)

The probability that the contractor will get both the plumbing contract and the electric contract is \( \frac{14}{45} \).

Summary of Probabilities

Conclusion

Based on the calculations using the given probabilities and the formula for the probability of the union of two events, the probability that the contractor gets both the plumbing and electric contracts is \( \frac{14}{45} \).

Revision Table: Key Probability Concepts

Additional Information: Understanding "At Least One" Probability

The phrase "at least one" in probability problems typically refers to the union of events. For two events, A and B, "at least one" means A happens OR B happens OR both happen. This is exactly what the union \( A \cup B \) represents.

The formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) is crucial here. When you simply add \(P(A)\) and \(P(B)\), you are double-counting the outcomes where both A and B occur. Therefore, you must subtract the probability of the intersection, \(P(A \cap B)\), once to correct this double-counting.

In this specific problem, we were given \(P(A \cup B)\) and asked to find \(P(A \cap B)\), which required rearranging the standard union formula. This is a common technique in probability questions.