13.4: More Rules for Calculating Probabilities

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Conjunctions with Independent Conjuncts

A conjunction is an and-sentence. The sentence, ‘Wilbur passed the final and Betty passed the final’ is a conjunction. The two simpler sentences glued together by the ‘and’ are called conjuncts (the order of conjuncts in a conjunction doesn’t matter). A conjunction is true just in case both of its conjuncts are true; if either conjunct is false, the whole thing is false. We will use ‘&’ to abbreviate ‘and’.

Independence

Two sentences are independent (of each other) just in case they are completely irrelevant to each other. The truth-value of one has no effect or influence or bearing on the truth value of the other. Knowing that one is true (or false) tells you nothing about whether the other is true (or false). Independence is a two-way street: if one thing is independent of a second, the second is also independent of the first.

Example 1: You are drawing cards from a deck, and after each draw you replace the card and reshuffle the deck. The results of the two draws are independent. What you get on the first draw has no influence on what you get on the second.

Example 2: You are drawing cards from a deck without replacing them. What you get on the first draw changes the makeup of the deck, and so the outcome of the first draw does bear on the outcome of the second. The outcome of the second draw is (to some degree) dependent on the outcome of the first.

Do not confuse incompatibility with independence. They are completely different.

Two things are incompatible just in case they cannot both be true at the same time; the truth of either excludes the truth of the other.
Two things are independent just in case the truth-value of each has no bearing on the truth value of the other.

Example: Getting a head on the next toss of a coin and a tail on that same toss are incompatible. But they are not independent.

Exercises

Which of the following pairs are incompatible? Which are independent?

Getting a 1 on the next roll of a die. Getting a 3 on that same roll.
Drawing an ace on the first draw from a deck. Drawing a jack on that same draw.
Getting a 1 on the next roll of a die. Getting a 3 on the roll after that.
The former host of Celebrity Apprentice is reelected President. I roll a 3 on the first roll of a die.
Getting a head on the next flip on a coin. Getting head on the flip after that.
Passing all of the exams in this course. Passing the course itself.

Rule for Conjunctions with Independent Conjuncts

Rule 5. (conjunctions with independent conjuncts): If the sentences A and B are independent, then the probability that their conjunction, A & B, is true, is Pr(A) times Pr(B).

Pr(A & B) = Pr(A) x Pr(B)

So, when two things are independent, the probability of their joint occurrence is determined by the simple multiplicative rule: multiply the probability of one by the probability of the other.

Example: What happens on the first toss of a coin has no effect on what happens on the second; getting a head on the first toss of a coin (H₁) and getting a head on the second toss (H₂) are independent. Hence, Pr(H₁ & H₂) = Pr(H₁) x Pr(H₂) = ½ x 1/2 (= ¼)

Screenshot (79).png — Figure \(\PageIndex{1}\): Tree Representation of the Probability of a Conjunction

The tree diagram (Figure 13.4.1) represents the possible outcomes. The numbers along each path represent the probabilities. The probability of a heads on the first flip (represented by the first node of the top path) is 1/2, and the probability of a second head (represented by the node at the upper right) is also 1/2. There are four paths through the tree, and each represents one possible outcome. Since all four paths are equally likely, the probability of going down any particular one is 1/4.

We can present the same information in a table (Figure 13.4.2) that shows more clearly why we multiply the probabilities of the two conjuncts. The outcomes along the side represent the two possible outcomes on the first toss, and the outcomes along the top represent the two outcomes of the second toss.

Screenshot (78).png — Figure \(\PageIndex{2}\): Table Representation of the Probability of a Conjunction

We can extend our rule to conjunctions with more than two conjuncts. Provided each conjunct is independent of all the rest, we can determine the probability of the entire conjunction by multiplying the individual probabilities of each of its conjuncts. For example, the probability that I will get heads on three successive flips of a coin is 1/2 x 1/2 x 1/2.

Our work will be much simpler because of the following facts.

Incompatibility is only relevant for disjunctions. We do not need to worry about whether the conjuncts of a conjunction are incompatible or not.
Independence is only relevant for conjunctions. We do not need to worry about whether the disjuncts of a disjunction are independent or not.

Winning the Lottery

The chances of winning a state lottery are very low; you have much better chances of winning in almost any casino in the world. To see why, imagine a lottery where you must correctly guess a one-digit number. There are 10 such digits, so your chances are 1 in 10, or .1. So far, so good. But now imagine that you must guess a two-digit number. There are ten possibilities for the first digit and ten possibilities of the second. Assuming the two digits are independent, this means that the chances of correctly guessing the first digit and the second digit are 1/10 x 1/10 = 1/100. You would win this lottery about once every 100 times you played. This may not sound so bad. But most state lotteries require you to match about twelve one-digit numbers. In this case, we determine the probability of winning by multiplying 1/10 by itself twelve times. When we write 1/10₁₂ out the long way, it turns out to be:

1 / 1,000,000,000

which is almost infinitesimally small.

Disjunctions with Compatible Disjuncts

Whenever the disjuncts of a disjunction are incompatible, R4 applies, but when they are compatible, we need a subtler rule.

It will help you to see why if we consider the following example. We are going to flip a quarter twice. What is the probability of getting a head on at least one of the two tosses; what is Pr(H₁ or H₂)? The probability of getting heads on any particular toss is 1/2. So, if we used our old disjunction rule (Rule 4., for incompatible disjuncts), we would have Pr(H₁ or H₂) + Pr(H₁) + Pr(H₂), which is just 1/2 + 1/2, or 1. This would mean that we were certain to get a head on at least one of our two tosses. But this is obviously incorrect, since it is quite possible to get two tails in a row.

Indeed, if we used our old disjunction rule to calculate the probability of getting a head on at least one of three tosses, we would have 1/2 + 1/2 + 1/2, which would give us a probability greater than 1.5 (and this could never be correct, since probabilities can never be greater than 1).

Screenshot (77).png — Figure \(\PageIndex{3}\): Disjunctions with Compatible (“Overlapping”) Disjuncts

If A and B are compatible, it is possible that they could occur together. For example, drawing an ace and drawing a black card are compatible (we might draw the ace of spades or the ace of clubs). We indicate this in Figure 13.4.3 by making the circle representing A and the circle representing B overlap. The overlapping, cross-hatched, region represents the cases where A and B overlap.

In terms of muddy diagrams, we add the weight of the mud on A to the weight of the mud on B, but when we do this, we weigh the mud where they overlap twice.

So, we must subtract once to undo this double counting. We must subtract the probability that A and B both occur, so that this area only gets counted once.

The General Disjunction Rule

Rule 6. (disjunctions): The probability of any disjunction, incompatible or compatible, is the sum of the probabilities of the two disjuncts, minus the probability that they both occur.

Pr(A or B) = Pr(A) + Pr(B) - Pr(A & B)

Example 1: Drawing an ace and drawing a club are not incompatible. So, Pr(A or C) = Pr(A) + Pr(C) - Pr(A & C); so it equals 1/13 + 1/4 - 1/52. We subtract the 1/52, because otherwise we would be counting the ace of clubs twice (once when we counted the aces, and a second time when we counted the clubs).

Example 2: Getting heads on the first and second flips of a coin are compatible. So to calculate Pr(H₁ or H₂), we have to subtract the probability that both conjuncts are true. We must consider Pr(H₁) + Pr(H₂) - Pr(H₁ & H₂), which is 1/2 + 1/2 - 1/4 (= 3/4).

Rule 6 is completely general; it applies to all disjunctions. But when the two disjuncts are incompatible, the probability that they are both true is 0, so we can forget about subtracting anything out.