The probability that a new ship will get an award for its design is $0.25$, the probability that it will get an award for the efficient use of materials is $0.35$, and that it will get both awards is $0.15$. What is the probability, that
- it will get at least one of the two awards;
- it will get only one of the awards.
Let $D$ be the event that a ship gets an award for its design.
Let $M$ be the event that a ship gets an award for the efficient use of materials.
Then, $\;$ $P \left(D\right) = 0.25$, $\;$ $P \left(M\right) = 0.35$ $\;$ and $\;$ $P \left(D \cap M\right) = 0.15$
- Let $A$ be the event that the ship gets at least one of the two awards
$\begin{aligned} P \left(A\right) & = P \left(D \;\; OR \;\; M\right) \\\\ & = P \left(D \cup M\right) \\\\ & = P \left(D\right) + P \left(M\right) - P \left(D \cap M\right) \\\\ & = 0.25 + 0.35 - 0.15 \\\\ & = 0.45 \end{aligned}$ - Let $B$ be the event that the ship gets only one of the awards
$\begin{aligned} P \left(B\right) & = P \left(\text{only D OR only M}\right) \\\\ & = P \left[\left(D \cap \overline{M}\right) \cup \left(\overline{D} \cap M\right)\right] \\\\ & = P \left(D \cap \overline{M}\right) + P \left(\overline{D} \cap M\right) \\\\ & = \left\{P \left(D\right) - P \left(D \cap M\right)\right\} + \left\{P \left(M\right) - P \left(D \cap M\right)\right\} \\\\ & = \left\{0.25 - 0.15\right\} + \left\{0.35 - 0.15\right\} \\\\ & = 0.10 + 0.20 \\\\ & = 0.30 \end{aligned}$
