I roll three different loaded dice. For the first die, the probability of getting a $5$ is $0.7$, for the second die the probability of getting a $5$ is $0.48$, and for the third die the probability of getting a $5$ is $0.38$.
What is the probability that the number $5$ comes up on exactly two of the three dice?
If $E_i$ is the event that dice $i$ ($i=1,2,3$) comes up a $5$, then the event of getting exactly two 5s is $$ \big(E_1 \cap E_2 \cap E_3^c\big)\cup\big(E_1 \cap E_2^c \cap E_3\big)\cup(E_1^c \cap E_2 \cap E_3\big) $$ Each of the pieces in parentheses is disjoint from the others (why!?) so using the properties of the probability $P$, $$ P(\text{exactly two } 5s) = P\big\{\big(E_1 \cap E_2 \cap E_3^c\big)\cup\big(E_1 \cap E_2^c \cap E_3\big)\cup(E_1^c \cap E_2 \cap E_3\big)\big\} = P\big(E_1 \cap E_2 \cap E_3^c\big) + P\big(E_1 \cap E_2^c \cap E_3\big)+P\big(E_1^c \cap E_2 \cap E_3\big). $$ Now, assuming that the $E_i$ are independent, $$ P\big(E_1 \cap E_2 \cap E_3^c\big) + P\big(E_1 \cap E_2^c \cap E_3\big)+P\big(E_1^c \cap E_2 \cap E_3\big) = P(E_1)P(E_2)P(E_3^c) + P(E_1)P(E_2^c)P(E_3) +P(E_1^c)P(E_2)P(E_3) $$ and now you can figure out the right side of the last equality with the information you're given.