Question

a construction company is planning to bid on a building contract. the bid costs the company $2000. the probability that the bid is accepted is \LXI0. if the bid is accepted, the company will make $26,000 minus the cost of the bid. \\(\dots\\) a. what is the expected value in this situation? $\square$ (round to the nearest dollar.)

Explanation:

Step1: Identify possible outcomes and their probabilities

• Outcome 1: Bid is accepted. Probability \( P_1 = \frac{1}{4} \). The profit here is \( 26000 - 2000 = 24000 \) dollars (since they get \( 26000 \) but spent \( 2000 \) on the bid).
• Outcome 2: Bid is rejected. Probability \( P_2 = 1 - \frac{1}{4}=\frac{3}{4} \). The profit here is \( - 2000 \) dollars (they spent \( 2000 \) on the bid and get nothing in return).

Step2: Calculate expected value

The formula for expected value \( E \) is \( E=\sum_{i}P_i\times V_i \), where \( V_i \) are the values of each outcome.

So, \( E = P_1\times V_1+P_2\times V_2 \)

Substitute the values:

\( E=\frac{1}{4}\times24000+\frac{3}{4}\times(- 2000) \)

First, calculate \( \frac{1}{4}\times24000 = 6000 \)

Second, calculate \( \frac{3}{4}\times(-2000)=-1500 \)

Then, \( E = 6000-1500=4500 \)

Answer:

\( 4500 \)