Section 1.8 - Conditional Probability

Let the trial T_≥2 be flipping 10 coins such that at least one comes up heads. What is the probability that an outcome of T_≥2 has at least two heads?

The requirement that at least one coin land heads-up is a condition,^?Don’t confuse the condition on any trial outcome (“contains at least one head”) with the predicate on the satisfactory outcomes (contains at least two heads). The condition is satisfied by every outcome in the sample space, while the predicate may or may not be satisfied by outcomes. and the resulting probability is known as known as a conditional probability. There’s at least three ways to find conditional probabilities: direct counting, indirect counting, and compliment probability.^?In fact, these three ways can solve any probability problem, not just conditional-probability problems.

Before getting to the solutions, it’s important to recall a fact, and understand why it’s important. What is T_≥2’s sample space? The sample space is the set of all possible trial outcomes. T_≥2’s sample space is not { t_i | 0 ≤ i < 1024 }^?The subscript i is interpreted as a 10-bit binary number with 1 bits interpreted as heads. because t₀ is not a possible outcome for T_≥2.^?t₀ contains no heads, which contradicts the condition that T_≥2 produce outcomes with at least one head. Instead, T_≥2’s sample space is { t_i | 0 < i < 1024 }, because now every t_i has at least one heads. The sample space is symmetric, and of size 1023, which means t_i for 0 < i < 1024 has probability 1/1023. These are the two most common errors when dealing with conditional probabilities:^?That is, these are the two mistakes I make the most. forgetting to adjust the sample space to accommodate the conditions, and misadjusting the sample space so it contains too many or too few outcomes.

The simplest and most straightforward solution (in some sense) is to directly count all the outcomes satisfying the predicate P(t_i) = t_i has at least two heads and add their probabilities together. There are 1013 such outcomes,^?This result courtesy of the haskell expression length [ i | i ← [1..1023],ones i > 1], where ones n returns the number of one bits in the binary representation of n. and the resulting probabilities sum to 1013/1023 = 0.99.

Direct counting doesn’t scale well with sample-space size or outcome predicate complexity. However, because the predicate partitions the sample space into two subsets, it may be possible to indirectly count the satisfactory outcomes by counting the unsatisfactory outcomes and using the fact that

unsatisfactory size + satisfactory size = sample-space size

to get

satisfactory size = sample-space size - unsatisfactory size

In this case, unsatisfactory size is 10,^?There are ten ways to toss ten coins so that exactly one heads comes up. and so satisfactory size = 1023 - 10 = 1013, and the probability is 1013/1023 = 0.99.

The third solution is a variant of the indirect counting because it too relies on outcome predicate partitioning:

Pr(unsatisfactory outcome) + Pr(satisfactory outcome) = Pr(sample-space outcome)

Pr(sample-space outcome) is certainty—that is, 1—and so

Pr(satisfactory outcome) = 1 - Pr(unsatisfactory outcome)

Pr(unsatisfactory outcome) is known as the compliment probability. In this case, Pr(unsatisfactory outcome) = 10/1023 = 0.01, to get Pr(satisfactory outcome) = 1 - 0.01 = 0.99.

Indirect counting and compliment probability are useful to the extent that counting the unsatisfactory outcomes is easier than counting the satisfactory outcomes. If counting the two kinds of outcomes are of similar difficultly, it’s likely that none of the three solution techniques is going to have a clear advantage over the others.


Example 1.8-2. Roll two fair die; T is the trial in which the outcome contains at least one even number. What is the probability that the outcome sums to 8?

T’s sample space is not { (x, y) | x, y ∈ [1..6] } because, for example, the sample space contains (3, 3), which is not an outcome for T.^?(3, 3) doesnt contain an even number.' T's sample space is

{ (x, y) | x, y ∈ [1..6] ∧ (even(x) ∨ even(y)) }

The haskell expression

length [ (x, y) | x <- [1..6], y <- [1..6], even x || even y ]

tells us the sample-space size is 27. The sample space is symmetric, and each outcome has probability 1/27.

Direct counting solution: How many outcomes sum to 8? Because every outcome contains an even number, and even + odd is odd (and 8 is even), even-odd rolls can be ignored. Of the even-even pairs, how many sum to 8?

D1	2	2	2	4	4	4	6	6	6
D2	2	4	6	2	4	6	2	4	6
sum	4	6	8	6	8	10	8	10	12

There are three satisfactory outcomes, and the probability of T producing one of the is 3/27 = 1/9.

Indirect-counting solution: How many outcomes do not sum to eight? All the pairs containing an odd number don't, and there are 9 + 9 = 18 of those.^?There are 3 possible values for the first coordinate, { 2, 4, 6 } and the same for the second { 1, 3, 5 } for a total of nine even-odd pairs. There are the same number of odd-even pairs for a total of 18 pairs. There may be a tendency to forget that not all the even-even pairs sum to eight,^?That is, I forgot the first time I solved this problem. and forget to count the ones that don't. From the previous solution, six even-even pairs don't sum to eight, for a total of 18 + 6 = 24 unsatisfactory outcomes. There are 27 - 24 = 3 satisfactory outcomes, for a probability of 3/27 = 1/9.

Compliment-probability solution: From the indirect-counting solution, there are 24 unsatisfactory outcomes, and the probability of producing one is 24/27 = 8/9. The probability of producing a satisfactory outcome is 1 - 8/9 = 1/9.

In this example it was more complicated to count the unsatisfactory outcomes than it was to count the satisfactory outcomes.