The problem describes a 911-response center that dispatches EMTs (E), firefighters (F), or police (C). We are given the percentages of calls requiring various combinations of these services. Part (a) asks for the probability that a random call requires exactly one of the three services. Part (b) asks for the probability that a random call requires all three services.

1. Problem Description

The problem describes a 911-response center that dispatches EMTs (E), firefighters (F), or police (C). We are given the percentages of calls requiring various combinations of these services. Part (a) asks for the probability that a random call requires exactly one of the three services. Part (b) asks for the probability that a random call requires all three services.

2. Solution Steps

(a) Let $P(E)$ be the probability that a call requires EMTs, $P(F)$ be the probability that a call requires firefighters, and $P(C)$ be the probability that a call requires police. We are given:

$P(E) = 0.40$

$P(F) = 0.41$

$P(E \cup F) = 0.64$

$P(E \cup C) = 0.92$

$P(F \cup C) = 0.77$

$P(E \cap F \cap C') = 0.06$, which means $P(E \cap F) - P(E \cap F \cap C) = 0.06$

$P(E' \cap F' \cap C') = 0.05$

We want to find the probability that exactly one service is required. Let $E_o$ be the event that only EMTs are required, $F_o$ be the event that only Firefighters are required, and $C_o$ be the event that only Police are required. We want to find $P(E_o) + P(F_o) + P(C_o)$.

Using the inclusion-exclusion principle:

$P(E \cup F) = P(E) + P(F) - P(E \cap F)$

$0.64 = 0.40 + 0.41 - P(E \cap F)$

$P(E \cap F) = 0.40 + 0.41 - 0.64 = 0.17$

Since $P(E \cap F) - P(E \cap F \cap C) = 0.06$, we have

$0.17 - P(E \cap F \cap C) = 0.06$

$P(E \cap F \cap C) = 0.17 - 0.06 = 0.11$

Now, $P(E \cup F \cup C) = 1 - P(E' \cap F' \cap C') = 1 - 0.05 = 0.95$

Also, $P(E \cup F \cup C) = P(E) + P(F) + P(C) - P(E \cap F) - P(E \cap C) - P(F \cap C) + P(E \cap F \cap C)$

$0.95 = 0.40 + 0.41 + P(C) - 0.17 - P(E \cap C) - P(F \cap C) + 0.11$

$P(E \cup C) = P(E) + P(C) - P(E \cap C)$

$0.92 = 0.40 + P(C) - P(E \cap C)$

$P(C) = 0.92 - 0.40 + P(E \cap C)$

$P(C) = 0.52 + P(E \cap C)$

$P(F \cup C) = P(F) + P(C) - P(F \cap C)$

$0.77 = 0.41 + P(C) - P(F \cap C)$

$P(C) = 0.77 - 0.41 + P(F \cap C)$

$P(C) = 0.36 + P(F \cap C)$

So, $0.95 = 0.40 + 0.41 + P(C) - 0.17 - P(E \cap C) - P(F \cap C) + 0.11$

$0.95 = 0.75 + P(C) - P(E \cap C) - P(F \cap C)$

$0.2 = P(C) - P(E \cap C) - P(F \cap C)$

Substitute $P(C) = 0.52 + P(E \cap C)$ and $P(C) = 0.36 + P(F \cap C)$

$0.2 = P(C) - (P(C) - 0.52) - (P(C) - 0.36)$

$0.2 = P(C) - P(C) + 0.52 - P(C) + 0.36$

$0.2 = 0.88 - P(C)$

$P(C) = 0.88 - 0.2 = 0.68$

$P(E \cap C) = P(C) - 0.52 = 0.68 - 0.52 = 0.16$

$P(F \cap C) = P(C) - 0.36 = 0.68 - 0.36 = 0.32$

$P(E_o) = P(E) - P(E \cap F) - P(E \cap C) + P(E \cap F \cap C) = 0.40 - 0.17 - 0.16 + 0.11 = 0.18$

$P(F_o) = P(F) - P(E \cap F) - P(F \cap C) + P(E \cap F \cap C) = 0.41 - 0.17 - 0.32 + 0.11 = 0.03$

$P(C_o) = P(C) - P(E \cap C) - P(F \cap C) + P(E \cap F \cap C) = 0.68 - 0.16 - 0.32 + 0.11 = 0.31$

$P(\text{exactly one}) = P(E_o) + P(F_o) + P(C_o) = 0.18 + 0.03 + 0.31 = 0.52$

(b) We found that $P(E \cap F \cap C) = 0.11$. This is the probability that a random call requires all three services.

3. Final Answer

a) $0.52$

b) $0.11$