Q.A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, $0.15, 0.20, 0.31, 0.26, .08$. Find the probabilities that a particular surgery will be rated

The key idea here is the Probability Addition Rule for Mutually Exclusive Events. When two events, $A$ and $B$, cannot occur at the same time (they are mutually exclusive), the probability that either $A$ or $B$ occurs is the sum of their individual probabilities.

The given probabilities for the types of surgeries are:

• $P(\text{Very Complex}) = 0.15$
• $P(\text{Complex}) = 0.20$
• $P(\text{Routine}) = 0.31$
• $P(\text{Simple}) = 0.26$
• $P(\text{Very Simple}) = 0.08$

We can now find the probabilities for each scenario:

(a) Complex or very complex:

These are mutually exclusive events.

$P(\text{Complex or Very Complex}) = P(\text{Complex}) + P(\text{Very Complex})$ …

When we talk about the probability of one event or another occurring, we often use the Addition Rule of Probability. This rule is particularly straightforward when the events are mutually exclusive.

Two events are mutually exclusive if they cannot happen at the same time. For instance, a surgery cannot be both "complex" and "routine" simultaneously. If you classify a surgery into one of these categories, it automatically excludes it from being in any other category. In this problem, the five categories of surgery complexity (very complex, complex, routine, simple, very simple) are all mutually exclusive.

Let's denote the probabilities for each rating:

• $P(VC) = P(\text{Very Complex}) = 0.15$
• $P(C) = P(\text{Complex}) = 0.20$
• $P(R) = P(\text{Routine}) = 0.31$
• $P(S) = P(\text{Simple}) = 0.26$
• $P(VS) = P(\text{Very Simple}) = 0.08$

First, it's a good practice to verify that the sum of all given probabilities is $1$:

$0.15 + 0.20 + 0.31 + 0.26 + 0.08 = 1.00$. This confirms that all possible outcomes are accounted for.

Now, let's solve each part of the problem.

1. Probability that a particular surgery will be rated complex or very complex.

   We are looking for $P(C \text{ or } VC)$. Since "complex" and "very complex" are mutually exclusive categories, we can simply add their probabilities.

   $P(C \text{ or } VC) = P(C) + P(VC)$

   $P(C \text{ or } VC) = 0.20 + 0.15$

   $P(C \text{ or } VC) = 0.35$

2. Probability that a particular surgery will be rated neither very complex nor very simple.

   This means the surgery must fall into one of the remaining categories: complex, routine, or simple.

   We can calculate this in two ways:

   • Method 1: Summing the probabilities of the desired outcomes. $P(\text{neither VC nor VS}) = P(C) + P(R) + P(S)$ $P(\text{neither VC nor VS}) = 0.20 + 0.31 + 0.26$ $P(\text{neither VC nor VS}) = 0.77$
   • Method 2: Using the complement rule. The event "neither very complex nor very simple" is the complement of the event "very complex or very simple". Let $A = \text{Very Complex}$ and $B = \text{Very Simple}$. $P(\text{neither A nor B}) = 1 - P(A \text{ or } B)$ …