## Analyzing Questions

- Mutually exclusive events cannot occur at the same time, while independent events do not influence each other's outcomes.
- The conditions of mutually exclusive and independent are more applicable to objects like dice, coins, and cards than to categorizing people due to the complexity and exceptions in human characteristics.
- Processes without replacement are never considered independent since the outcome of one event influences the probability of subsequent events.
- A detailed example problem demonstrates how to apply these concepts by dividing the problem into stages and analyzing the independence and mutual exclusivity of events, as well as using the complement rule for simplification.
- The example problem, involving coin tosses and dice rolls, illustrates the application of probability rules to calculate the likelihood of winning a game.

**Q: **What is the difference between "mutually exclusive" and "independent"?

**A: "**Mutually exclusive" means that A and B cannot both occur; P(Both A and B) = 0. There is no way for both A and B to happen. If A happens, then B cannot happen, and if B happens, A cannot happen.

Example: flipping a coin once, getting a head (A) and getting a tails (B) would be mutually exclusive events. They cannot both occur.

Now let's talk about independent events:

"Independent" means that whether A occurs has no effect on whether B occurs, and vice-versa.

If two events are independent, then P(A and B) = P(A)P(B) They can both occur, and the probability of both occurring is the product of their individual probabilities.

Example: flipping a coin twice. The probability of getting a head on the first flip (X) and no effect on getting a head on the second flip (Y). They are independent events. The probability of getting two heads is: P(X)P(Y)

So, to repeat:

If two events A and B are independent, then that means:

P(Both A and B) = P(A)*P(B)

Whether A occurs does not affect the probability of whether B occurs, and vice-versa.

If A and B are mutually exclusive, then that means:

P(Both A and B) = 0

If A occurs, then B cannot occur also, and vice versa.

**Q: **How is 7/8 the probability of advancing to phase 2? It seems like we're counting three tosses even after we encounter first head?

**A: **Events are independent if the probability of their *outcomes* are not affected by each other. The fact that we flip a heads first does not change the probability of flipping a heads a the 2nd time. True, we don' *need* to flip the 2nd time if we flip a heads first, but the 2nd flip, if we did it, would still be independent.

Even if we flip heads on the first toss, we *could* still flip the coin the 2nd and 3rd times. All 4 outcomes would be valid since we already flipped a heads on the first flip.

HHH Probability = (1/2)^3 = 1/8

HHT Probability = 1/8

HTT Probability = 1/8

HTH Probability = 1/8

Notice 4 * (1/8) = 1/2, which is the probability of getting heads on the first flip.

Let's look at it another way:

probability of winning first phase =

P(winning on 1st flip or winning 2nd flip or on 3rd flip) =

P(winning on 1st flip) + P(winning on 2nd flip) + P(winning on 3rd flip)

P(winning on 1st flip) = 1/2

P(winning on 2nd flip) = P(Tails on 1st flip) * P(Heads on 2nd flip) = 1/2 * 1/2 = 1/4

P(winning on 3rd flip) = P(tails on 1st flip) + P(tails on 2nd flip) * P(heads on 3rd flip) = 1/2 * 1/2 * 1/2 = 1/8

So the probability of success in the first phase is:

1/2 + 1/4 + 1/8 = 7/8

There are 8 equally likely outcomes:

HHH

HHT <--- 4/8 with heads on 1st flip

HTT

HTH

THH <--- 2/8 with heads on 2nd flip

THT

TTH <--- 1/8 with heads on 3rd flip

TTT <--- 1/8 with no heads

So again, we have:

P(heads on 1st flip) + P(heads on 2nd flip) + P(heads on 3rd flip) =

1/2 + (1/2)(1/2) + (1/2)(1/2)(1/2) =

1/2 + 1/4 + 1/8 =

7/8

In the same way, whether or not we advance to the second phase depends on our success in the first phase. But, *once we get to the 2nd phase,* the probability of success on the 2nd phase is independent of what happened in the 1st phase.