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Lesson by Mike McGarry
Magoosh Expert

Summary
The content provides an in-depth exploration of applying mutually exclusive and independent rules in probability questions, specifically within the context of GMAT preparation. It emphasizes understanding these concepts through practical examples and solving a complex probability problem by breaking it down into simpler, independent stages.
• Mutually exclusive events cannot occur simultaneously, exemplified by drawing cards or rolling dice.
• Independent events are those where the outcome of one event does not affect the outcome of another, a concept crucial for solving probability problems.
• The conditions of mutually exclusive and independent are less applicable to categories involving people due to the complexity and exceptions within human characteristics.
• Processes without replacement are never independent, affecting the probability calculations in sequential events.
• Solving complex probability problems involves breaking down the problem into stages, analyzing independence, and applying the complement rule to simplify calculations.
Chapters
00:00
Understanding Mutually Exclusive and Independent Events
01:35
Applicability of Probability Rules
02:56
The Impact of Without Replacement
04:05
Solving a Complex Probability Problem

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: