## Analyzing Questions

- Mutually exclusive events cannot occur at the same time, while independent events do not influence each other's outcomes.
- The categorization of events as mutually exclusive or independent is more straightforward with inanimate objects like dice and cards than with complex categories like human demographics.
- Processes involving 'without replacement' are never considered independent since the outcome of one event influences the next.
- A detailed example problem involving coin flips and dice rolls illustrates how to apply these concepts by dividing the problem into independent stages and using the complement rule.
- The example problem demonstrates a higher difficulty level than typically found on the GRE but serves to reinforce understanding of probability concepts.

**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.