The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that
(ii) it will get only one of the awards.

Answer

To solve the problem, we need to find the probability that the railway bridge will get only one of the two awards: either for design or for efficient use of materials, but not both.

Let's denote:
- \( P(D) \) = Probability of winning an award for design = 0.48
- \( P(M) \) = Probability of winning an award for efficient use of materials = 0.36
- \( P(D \cap M) \) = Probability of winning both awards = 0.2

We want to find the probability that the bridge gets only one of the awards, which can be expressed as:
\[ P(\text{only } D) + P(\text{only } M) \]
This can be rewritten using the intersection and complement notation:
\[ P(D \cap M') + P(M \cap D') \]

Using the properties of probabilities, we can express these as:
\[ P(D \cap M') = P(D) - P(D \cap M) \]
\[ P(M \cap D') = P(M) - P(D \cap M) \]

Now, substituting the values we have:
1. Calculate \( P(D \cap M') \):
\[ P(D \cap M') = P(D) - P(D \cap M) = 0.48 - 0.2 = 0.28 \]

2. Calculate \( P(M \cap D') \):
\[ P(M \cap D') = P(M) - P(D \cap M) = 0.36 - 0.2 = 0.16 \]

3. Now, add these two probabilities together to find the probability of getting only one award:
\[ P(\text{only one award}) = P(D \cap M') + P(M \cap D') = 0.28 + 0.16 = 0.44 \]

Thus, the probability that the railway bridge will get only one of the awards is **0.44**.