Probability of two shooters A and B that are shooting separately targets

Two shooters A and B are shooting two separately targets, Probabilities of target hitting are 0.3 For A and 0.6 For B. Calculate these probabilities :

A) At least one shooter hits the target:

My answer : P(A) + P(B) - P(A)*P(B) = 0.3 + 0.6 - 0.18 = 0.72

B) Just only one shooter hit the target:

My Answer: P(A-B) + P(B-A) = P(A) + P(B) - 2[P(A)*P(B)] = 0.3 + 0.6 - 2*0.18 = 0.54

C) None of them shootings hit the target:

My Answer : 1 - P(A) + P(B) - P(A)*P(B) = 1 - 0.72 = 0.28

Are my answers correct? Any idea to solve these problems in another ways ?

• Another way for the last one: $P(\overline{A \ or\ B}) =P(\bar{A} \ and \ \bar{B}) = (0.7)(0.4)$ Oct 20, 2018 at 4:05

For part $$b$$, we can make use of part $$(a)$$, just subtract $$P(A \cap B)$$ directly.
For part $$c$$, while the final answer is correct, remember to use braces.
$$1 - \color{blue}(P(A) + P(B) - P(A)P(B)\color{blue})$$