Answer:
The probability distribution of winning x bids in the scenario where a construction company has a 20% chance of winning any given bid can be described using the binomial distribution. The binomial distribution is used to model the number of successes in a fixed number of independent trials, where each trial has the same probability of success.
In this case, if the company is considering submitting bids for two contracts, the probability distribution of winning x bids can be calculated as follows:
Let's denote:
- n as the number of trials (in this case, the number of contracts the company is considering, which is 2)
- p as the probability of success on any given trial (in this case, the probability of winning a bid, which is 0.20)
- x as the number of successful bids
The probability mass function of the binomial distribution is given by:
\[ P(X = x) = \binom{n}{x} \times p^x \times (1 - p)^{n-x} \]
Where:
- \( \binom{n}{x} \) is the binomial coefficient, equal to n! / (x! * (n-x)!)
- \( p^x \) is the probability of getting x successes
- \( (1 - p)^{n-x} \) is the probability of getting n-x failures
For x = 0, 1, 2, the probability distribution of winning x bids can be calculated as follows:
For x = 0:
\[ P(X = 0) = \binom{2}{0} \times 0.20^0 \times (1 - 0.20)^{2-0} = 1 \times 1 \times 0.80^2 = 0.64 \]
For x = 1:
\[ P(X = 1) = \binom{2}{1} \times 0.20^1 \times (1 - 0.20)^{2-1} = 2 \times 0.20 \times 0.80 = 0.32 \]
For x = 2:
\[ P(X = 2) = \binom{2}{2} \times 0.20^2 \times (1 - 0.20)^{2-2} = 1 \times 0.20^2 \times 1 = 0.04 \]
So, the probability distribution of winning 0, 1, or 2 bids is:
- P(X = 0) = 0.64
- P(X = 1) = 0.32
- P(X = 2) = 0.04