4.4 Probability: Rules, Compound Events, and Conditional Probability

4.4 Probability: Rules, Compound Events, and Conditional Probability

Probability items reward one habit above all others: naming the structure before touching the numbers. Is this an or question or an and question? Are the events independent or does the first draw change the second? Is the question really asking for the complement in disguise? Get the structure right and the arithmetic on an entrance test is almost always a two-fraction multiplication you can do in your head.

1. Sample Space and the Basic Ratio

The sample space $S$ is the set of all possible outcomes; an event $E$ is any subset of it. When the outcomes are equally likely,

$P(E) = \frac{n(E)}{n(S)}, \qquad 0 \le P(E) \le 1$

A probability of 0 means impossible and 1 means certain. Any answer outside that interval — or any "probability" bigger than 1 produced by adding fractions — is an arithmetic error, not a result.

For small experiments, simply list $S$. Tossing three coins gives eight equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Exactly two heads occurs in three of them, so $P = 3/8$. Rolling two dice gives 36 ordered pairs, of which six sum to 7, so $P(\text{sum} = 7) = 6/36 = 1/6$.

Complement rule. The probability that $E$ does not happen is

$P(E') = 1 - P(E)$

2. The Addition Rule: "Or" Questions

Two events are mutually exclusive if they cannot both happen in one trial (drawing a card that is both a king and a queen). They overlap if they can (a card that is both a king and a heart). The general rule is

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

and the subtraction vanishes only in the mutually exclusive case. From a standard 52-card deck, $P(\text{king or heart}) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13$; forgetting to subtract the king of hearts double-counts one card. By contrast $P(\text{king or queen}) = 4/52 + 4/52 = 2/13$, since no card is both.

3. The Multiplication Rule: "And" Questions

Replacement is the switch. A bag holds 4 red and 6 blue marbles. Drawing two with replacement keeps the pool at 10 every time, so the draws are independent: $P(\text{both red}) = (4/10)^2 = 4/25$. Drawing two without replacement shrinks the pool: $P(\text{both red}) = (4/10)(3/9) = 2/15$, a smaller number, as it should be.

Note that mutually exclusive events with non-zero probabilities are never independent — if one happens, the other's probability collapses to zero. Treating those two words as synonyms is the single most expensive vocabulary error in this topic.

4. Conditional Probability and Two-Way Tables

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) > 0$

In words: the condition becomes the new denominator. A two-way table makes this visible, because conditioning simply means restricting yourself to one row or one column.

Worked Example 1 — a two-way table

Problem. Two hundred Grade 12 students at a Manila senior high school sat a mock entrance test. Their results by strand:

One student is chosen at random.

1. $P(\text{passed})$. Use the column total over the grand total: $135/200 = 27/40 = 0.675$.
2. $P(\text{STEM and passed})$. Read the single cell: $72/200 = 9/25 = 0.36$.
3. $P(\text{STEM or passed})$. Apply the addition rule: $\frac{100}{200} + \frac{135}{200} - \frac{72}{200} = \frac{163}{200} = 0.815$. Skipping the subtraction gives 235/200, which is greater than 1 and therefore visibly impossible.
4. $P(\text{passed} \mid \text{STEM})$. The condition restricts you to the STEM row, so the denominator is 100: $72/100 = 0.72$.
5. $P(\text{STEM} \mid \text{passed})$. Now the condition restricts you to the Passed column, so the denominator is 135: $72/135 = 8/15 \approx 0.533$.

Steps 4 and 5 use the same cell and give different answers. That is the whole lesson of conditional probability: $P(A \mid B)$ and $P(B \mid A)$ are different questions. Note also that $P(\text{passed} \mid \text{STEM}) = 0.72 \ne P(\text{passed}) = 0.675$, so strand and outcome are not independent here.

5. Without Replacement, Tree Diagrams, and "At Least One"

Worked Example 2 — drawing without replacement

Problem. A tray at a Quiapo bakery holds 12 pieces of pandesal: 5 with cheese filling and 7 plain. Aling Rosa picks two at random without looking and without putting the first one back. Find (a) $P(\text{both cheese})$, (b) $P(\text{exactly one cheese})$, and (c) $P(\text{at least one cheese})$.

                       cheese (4/11)  ->  C, C  =  5/12 x 4/11 = 20/132
         cheese (5/12)
                       plain  (7/11)  ->  C, P  =  5/12 x 7/11 = 35/132
start
                       cheese (5/11)  ->  P, C  =  7/12 x 5/11 = 35/132
         plain  (7/12)
                       plain  (6/11)  ->  P, P  =  7/12 x 6/11 = 42/132

(a) Both cheese. After one cheese roll is taken, 4 cheese pieces remain among 11: $P = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$

(b) Exactly one cheese. Two branches of the tree qualify — cheese then plain, and plain then cheese — and each has probability $35/132$: $P = 2 \times \frac{35}{132} = \frac{70}{132} = \frac{35}{66}$

(c) At least one cheese. Do not add three branches. Use the complement, because "at least one" is the opposite of "none": $P = 1 - P(\text{both plain}) = 1 - \frac{7}{12} \times \frac{6}{11} = 1 - \frac{42}{132} = \frac{90}{132} = \frac{15}{22}$

Check. Parts (a) and (b) should add to part (c): $\frac{5}{33} + \frac{35}{66} = \frac{10}{66} + \frac{35}{66} = \frac{45}{66} = \frac{15}{22}$. They do, and all four tree branches sum to 1.

6. Odds Versus Probability

Odds in favour compare favourable outcomes to unfavourable ones, not to the total.

• If $P(E) = a/(a+b)$, then the odds in favour of $E$ are $a : b$, and the odds against are $b : a$.
• Conversely, odds of $a : b$ in favour give $P(E) = \dfrac{a}{a+b}$.

So odds of $3 : 1$ mean a probability of $3/4$, not $1/3$. A probability of 0.6 corresponds to odds of $3 : 2$. The pandesal result $P = 15/22$ has 15 favourable parts and $22 - 15 = 7$ unfavourable ones, so the odds in favour are $15 : 7$.

7. Expected Value

The expected value of a numerical outcome is the sum of each value times its probability — the long-run average per trial, not a value you necessarily ever see.

$E = \sum x_i P(x_i)$

A fair die has $E = (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5$ pips per roll, even though 3.5 never appears. At a school fair, 100 raffle tickets are sold at ₱20 each; one ticket wins ₱500 and four tickets win ₱100. The expected prize per ticket is

$E = \frac{1}{100}(500) + \frac{4}{100}(100) + \frac{95}{100}(0) = 5 + 4 = \text{PHP } 9$

so the expected net result per ticket is $9 - 20 = -11\text{ PHP}$: a loss of about eleven pesos per ticket on average, which is precisely how the fair funds itself.

8. Common Traps

1. Adding overlapping probabilities. Any answer above 1 is a signal you forgot to subtract $P(A \cap B)$.
2. Treating a without-replacement draw as independent. The second denominator must drop by one.
3. Reversing a conditional. "Given that she passed" fixes the passed denominator, not the strand denominator.
4. Adding branches for "at least one". Use $1 - P(\text{none})$; it is faster and safer.
5. Reading odds as probability. Odds of $2 : 3$ are a probability of $2/5$.
6. Confusing mutually exclusive with independent. Mutually exclusive events are maximally dependent.
7. Ignoring order in "exactly one" problems. Two arrangements exist, so the single-branch probability must be doubled.