Mutually Exclusive vs. Independent - Extra Practice

Quick refresher

• Mutually exclusive: \(A\) and \(B\) can’t both happen. \(P(A \cap B) = 0\).
• Independent: knowing whether \(A\) happened doesn’t change the chance of \(B\). Check: does \(P(A \cap B) = P(A) \times P(B)\)?

These sound similar, but they’re nearly opposites. If \(A\) and \(B\) are mutually exclusive, then knowing \(A\) happened tells you a lot about \(B\): it definitely didn’t happen.

Question 1

Barnaby the parrot knows exactly three phrases: “Hello!”, “Pretty bird!” and “Where’s my cracker?”. Each time someone walks into the room, he squawks exactly one of them, chosen at random. Someone walks in.

• \(A\): Barnaby says “Hello!”
• \(B\): Barnaby says “Pretty bird!”

1. Are \(A\) and \(B\) mutually exclusive?

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Yes. Barnaby only says one phrase per visit, so he can’t say both. \(P(A \cap B) = 0\).

2. Are \(A\) and \(B\) independent?

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No. If you know Barnaby said “Hello!”, you know for certain he didn’t say “Pretty bird!”. Learning about \(A\) changed the chance of \(B\) from \(1/3\) all the way down to 0, so they’re dependent.

Question 2

Two rival wizards settle every argument the same way: each one flips their own fair coin, at the same time.

• \(A\): the first wizard’s coin lands heads
• \(B\): the second wizard’s coin lands heads

1. Are \(A\) and \(B\) mutually exclusive?

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No. Both coins can land heads, so \(A\) and \(B\) can happen together.

2. Are \(A\) and \(B\) independent?

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Yes. The coins don’t affect each other, no matter how much the wizards glare. \(P(A \cap B) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2} = P(A) \times P(B)\).

Question 3

You roll a fair six-sided die once.

• \(A\): you roll an even number
• \(B\): you roll a 4 or higher

1. Are \(A\) and \(B\) mutually exclusive?

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No. A 4 or a 6 makes both happen.

2. Are \(A\) and \(B\) independent?

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No. \(P(A) = \frac{3}{6}\) and \(P(B) = \frac{3}{6}\), so \(P(A) \times P(B) = \frac{1}{4}\). But \(A \cap B = \{4, 6\}\), so \(P(A \cap B) = \frac{2}{6} = \frac{1}{3}\). Since \(\frac{1}{3} \neq \frac{1}{4}\), they’re dependent: knowing the roll is even makes “4 or higher” more likely (\(\frac{2}{3}\) instead of \(\frac{1}{2}\)).

Question 4

You roll a fair six-sided die once. (Same die, different events.)

• \(A\): you roll an even number
• \(B\): you roll a 1 or a 2

1. Are \(A\) and \(B\) mutually exclusive?

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No. Rolling a 2 makes both happen.

2. Are \(A\) and \(B\) independent?

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Yes, surprisingly!

Independence doesn’t need two unrelated experiments. Two events about the same experiment can be independent, if the numbers work out.

The fast way: check if \(P(A) \times P(B) = P(A \cap B)\). Starting with the full die possible \(\{1, 2, 3, 4, 5, 6\}\):

• \(P(A) = 3/6 = 1/2\) and \(P(B) = 2/6 = 1/3\), so \(P(A) \times P(B) = 1/6\)
• \(A \cap B = \{2\}\), so \(P(A \cap B) = 1/6\). They match.

The longer, but clearer way: see if \(P(B)\) changes when \(A\) does (or doesn’t) happen. If it doesn’t change, they’re independent. Lets check:

• Out of the 6 equally-likely outcomes \(\{1, 2, 3, 4, 5, 6\}\), exactly two are a 1 or 2 (event \(B\)). Thus, \(P(B) = 2/6 = 1/3\) to start.
• If you know that you got an even number (\(A\) happened), this means that the roll had to be one of \(\{2, 4, 6\}\). Only 1 out of these 3 is a “1 or 2” (event \(A\)), so \(P(A) = 1/3\) still. Unchanged.
• Likewise, if you know that you got an odd number (\(A\) didn’t happen), this means that the roll had to be one of \(\{1, 3, 5\}\). Again, only 1 out of these is a “1 or 2” (event \(A\)), so \(P(A) = 1/3\). Unchanged again.
• In other words, whether the roll even or odd, 1/3 of the time it’s going to be a 1 or a 2. The events are independent.
• Notationally, we effectively just checked if \(P(A) = P(A \mid B) = P(A \mid B^C)\), which says the same thing as sentence above.

Question 5

You draw one card from a well-shuffled standard 52-card deck.

• \(A\): the card is a king
• \(B\): the card is a heart

1. Are \(A\) and \(B\) mutually exclusive?

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No. The king of hearts is both.

2. Are \(A\) and \(B\) independent?

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Yes.

• \(P(A) = 4/52 = 1/13\) and \(P(B) = 13/52 = 1/4\), so \(P(A) \times P(B) = \frac{1 \times 13}{13 \times 51} = \frac{1}{52}\).
• And the only card that fits both events is the king of hearts, so \(P(A \cap B) = \frac{1}{52}\). They match.
• Each suit has exactly one king, so knowing the suit tells you nothing about whether it’s a king, and vice versa.

Question 6

A gumball machine holds 3 red and 7 green gumballs. You buy a gumball, look at it, and (please don’t do this) put it back in the machine. Then you buy another one.

• \(A\): the first gumball is red
• \(B\): the second gumball is red

1. Are \(A\) and \(B\) mutually exclusive?

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No. Both gumballs can be red.

2. Are \(A\) and \(B\) independent?

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Yes. Because the first gumball goes back, the machine is exactly the same for the second purchase, so \(P(B) = \frac{3}{10}\) whether or not \(A\) happened. Drawing with replacement makes the draws independent.

Question 7

Same gumball machine: 3 red and 7 green. This time you’re a normal person, so you eat the first gumball before buying the second.

• \(A\): the first gumball is red
• \(B\): the second gumball is red

1. Are \(A\) and \(B\) mutually exclusive?

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No. Both gumballs can still be red.

2. Are \(A\) and \(B\) independent?

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No. If the first one was red, the machine now holds 2 red out of 9, so \(P(B \mid A) = \frac{2}{9}\). If the first one was green, it holds 3 red out of 9, so \(P(B \mid A^C) = \frac{3}{9}\). The chance of \(B\) depends on what happened first. Drawing without replacement makes the draws dependent.

Question 8

At a dog park, 60 dogs are surveyed (with their owners’ help). 18 dogs are wearing sweaters, 20 dogs are corgis,and 6 are corgis wearing sweaters. Pick one of these dogs at random.

• \(A\): the dog is a corgi
• \(B\): the dog is wearing a sweater

1. Are \(A\) and \(B\) mutually exclusive?

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No. 6 dogs are corgis in sweaters, so \(P(A \cap B) = \frac{6}{60} = 0.1\), which isn’t 0.

2. Are \(A\) and \(B\) independent?

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Yes.

Let’s do this one intuitively. A third of the dogs (20/60) are corgis. Also, a third of sweater-wearing dogs (6/18) are corgis. So knowing that a dog has a sweater doesn’t change its chance of being a corgi. They’re indpendent.

Question 9

A survey of 100 students finds that 30 have an 8am class, 40 students drink 3 or more coffees a day, and 20 do both. Pick a student at random.

• \(A\): the student has an 8 am class
• \(B\): the student drinks 3 or more coffees a day

1. Are \(A\) and \(B\) mutually exclusive?

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No. 20 students do both, so obviously both can happen.

2. Are \(A\) and \(B\) independent?

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No.

• \(P(A) \times P(B) = (30/100) \times (40/100) = 0.3 \times 0.4 = 0.12\).
• But we were told that 20 students do both, so \(P(A \cap B) = \frac{20}{100} = 0.2\).

These are not equal, so they’re not independent. In other words, knowing that someone has an 8am changes the odds that they drink 3 or more coffees a day. Students’ likely coffee consumption depends on whether they have an 8am class. (No one is surprised.)

Question 10

You draw two cards at the same time from a well-shuffled standard deck.

• \(A\): the first card is an ace
• \(B\): the second card is an ace

1. Are \(A\) and \(B\) mutually exclusive?

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No. Both cards can be aces (there are 4 aces in a deck).

2. Are \(A\) and \(B\) independent?

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No.

Think of it this way: you take two cards, but don’t look at them yet. At the moment, each card, as far as you know, could be any one of the 52 cards in the deck. 4 of those are aces, so \(P(A) = \frac{4}{52}\) and \(P(B) = \frac{4}{52}\) as well, from what you know so far.

But after you look at one of the cards, you now know that the other one can’t be that card.

• Let’s say you peek at the first card, and see the ace of clubs (A♣). Now you know that the second card can’t be the A♣: it can only be one of the remaining 51 cards. Of those, only 3 are aces (A♥︎, A♦︎, and A♠), so \(P(B \mid A) = \frac{3}{51}\). This is different from the original \(P(B) = \frac{4}{52}\), so these events are dependent.
• Even if the first card hadn’t been an ace, maybe the Jack of Hearts, you now know the second card can’t be that jack. But 4 aces are still out there, so \(P(B \mid A) = \frac{4}{51}\). This is still different from the original \(\frac{4}{52}\), so again the events are dependent.

Question 11

Events \(A\) and \(B\) have \(P(A) = 0.3\) and \(P(B) = 0.5\).

1. If \(A\) and \(B\) are mutually exclusive, what are \(P(A \cap B)\) and \(P(A \cup B)\)?

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\(P(A \cap B) = 0\). They can’t both happen (that’s the definition of mutually exclusive!)

But if they’re mutually exclusive, we can just add their probabilities (we don’t have to worry about double-counting their intersection, since it’s zero). So \(P(A \cup B) = P(A) + P(B) = 0.3 + 0.5 = 0.8\)

2. If instead \(A\) and \(B\) are independent, what are \(P(A \cap B)\) and \(P(A \cup B)\)?

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If they are independent, we can just multiply their probabilities. We used this very fact to test for independence earlier. So, \(P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.5 = 0.15\).

But since they are independent, they are not mutually exclusive! So we can’t just add their probabilities, as that will double-count the times when both happen. So we undo that double-count like so:

\(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.5 - 0.15 = 0.65\)