$$ 1 \leftarrow \quad $$ Find the probability $$ P(E $$ or $$ F) $$ if $$ E $$ and $$ F $$ are mutually exclusive, $$ P(E)=0.41 $$, and $$ P(F)=0.45 $$. The probability $$ P(E $$ or $$ F) $$ is $$ \square $$. (Simplify your answer.)

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Given that events $$ E $$ and $$ F $$ are mutually exclusive, they cannot occur at the same time. Therefore, the probability of either $$ E $$ or $$ F $$ occurring is given by the sum of their individual probabilities:

$$
P(E \text{ or } F) = P(E) + P(F)
$$

Substitute the given values:

$$
P(E \text{ or } F) = 0.41 + 0.45
$$

Simplify the sum:

$$
P(E \text{ or } F) = 0.86
$$

Thus, the probability $$ P(E \text{ or } F) $$ is $$ 0.86 $$.
The probability $$ P(E $$ or $$ F) $$ is $$ 0.86 $$.

Find the probability or if and are mutually exclusive, , and .
The probability or is . (Simplify your answer.)

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Find the probability or if and are mutually exclusive, , and . The probability or is . (Simplify your answer.)