Probability

Probability of dice outcomes

A fair die has six equally likely faces, so tossing n dice gives 6^n equally likely ordered outcomes, and the dice are independent. The standard method is to count the favourable ordered outcomes and divide by 6^n. When each die must lie in a fixed set S, the probability is \left(\frac{|S|}{6}\right)^n, because independence lets the factors multiply. For example, if each die must show a multiple of 3, that is 3 or 6 on every die, the probability for three dice is \left(\frac{1}{3}\right)^3.

When a question specifies the colour or class of each toss in a particular order, the probability is the product of the single-toss probabilities taken in that order, so the sequence matters. For classification questions, first group the faces into classes. On a die the primes are \{2,3,5\}, the composites are \{4,6\}, and 1 is neither prime nor composite. If three dice must show one face from each class, the favourable count is the number of class assignments multiplied by the 3! = 6 arrangements of the dice.

Sum-based questions are handled by counting the ordered tuples with a given total. Replacing each face x by 7-x is a bijection, so for three dice the number of ways to obtain sum s equals the number of ways to obtain 21-s. This symmetry reduces the work. A condition such as 'sum at least 15' is handled by adding the counts for totals 15 to 18. Perfect-square totals are found the same way, by counting each square separately and adding.

For two dice, a difference d > 0 arises from ordered pairs (a,b) with |a-b| = d. Each pair with distinct entries gives two ordered outcomes, so both orders must be counted before dividing by 36. For three distinct faces in arithmetic progression, read the dice as (x,y,z); the condition is y-x = z-y = d \neq 0, and counting the valid pairs (x,d) gives the favourable total.

Bounds of event probabilities

Every probability lies in [0,1], and the addition rule P(A\cup B) = P(A)+P(B)-P(A\cap B) turns this fact into bounds on the joint probability. Since P(A\cup B) \leq 1, we get P(A\cap B) \geq P(A)+P(B)-1. Also P(A\cap B) \geq 0, so the lower bound is \max\{0,\,P(A)+P(B)-1\}. Since A\cap B \subseteq A and A\cap B \subseteq B, we have P(A\cap B) \leq \min\{P(A),P(B)\}, and since A\cup B contains both events, P(A\cup B) \geq \max\{P(A),P(B)\}. These extremes are attained when one event is contained in the other.

The sum P(A)+P(B) equals P(A\cup B)+P(A\cap B), so its extreme values come from the extreme values of the union and intersection. The minimum uses the lower bound of the union and of the intersection together. The maximum uses the upper bound P(A\cup B)\leq 1 together with the largest possible intersection.

For three events, inclusion-exclusion gives P(A\cup B\cup C) = \sum P(\cdot) - \sum P(\cdot\cap\cdot) + P(A\cap B\cap C). If the union is bounded below and the pairwise intersections are known, the expression yields an upper bound on the pair P(B\cap C). The lower bound comes from the fact that B\cap C contains A\cap B\cap C.

A conditional probability also has bounds. From P(A\cap B)\geq P(A)+P(B)-1 and P(A\mid B)=P(A\cap B)/P(B), we get P(A\mid B)\geq \dfrac{P(A)+P(B)-1}{P(B)} for P(B)>0.

Probability of union of events

The union E\cup F means that at least one of the events occurs, and its probability is found with the addition rule P(E\cup F)=P(E)+P(F)-P(E\cap F). The subtraction removes the overlap that was counted twice. If E and F are mutually exclusive, P(E\cap F)=0 and the rule reduces to a plain sum. For a coin tossed twice, 'head on the first or the second toss' has probability 1 minus the probability of two tails, giving \frac{3}{4} on the four equally likely outcomes.

Useful relations between two events are: P(E\cup F)\leq P(E)+P(F), which follows because the overlap is non-negative, and P(E\cap F)\geq P(E)+P(F)-1, which follows because the union cannot exceed 1. A common error is to add the intersection term instead of subtracting it, which is never correct.

For independent events, the complement of the union is the intersection of the complements, so P(A\cup B) = 1 - P(\overline{A})\,P(\overline{B}). This form is often faster than the addition rule.

With given single and joint probabilities, the addition rule is applied directly. For instance, if the probability of owning item L is 0.68, of owning item D is 0.56, and of owning both is 0.48, then the probability of owning at least one is 0.68+0.56-0.48. The same rule can be rearranged to find a missing single probability, P(E) = P(E\cup P) - P(P) + P(P\cap E), when the other three quantities are given. The formula is used even when the given data are not fully consistent with a real experiment.

Conditional probability relations

The conditional probability of A given B is P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}, provided P(B) > 0. Rearranging gives the multiplication rule P(A\cap B) = P(A\mid B)\,P(B). Using this rule, a joint probability is found from one conditional and one marginal probability, and a second conditional probability then follows by dividing by the other marginal.

A key relation is that P(A\mid B) < P(A) holds exactly when P(A\cap B) < P(A)P(B), and this is the same condition as P(B\mid A) < P(B). Thus the two events are negatively associated in both directions, and the inequality does not need to be checked twice.

In a two-way table, a joint probability uses the total number of outcomes as the denominator, whereas a conditional probability uses the number of outcomes in the conditioning group as the denominator. For example, the probability of a graduate with less than the threshold experience is the joint count over the total, while the probability of a graduate given the same experience condition uses the smaller group. When the condition is that the applicant has less than three years of experience, the sample space is restricted to that group, and the probability is the count of graduates in that group divided by the group size. Similarly, for the group of non-graduates, the probability that they have less than three years of experience is the count in that cell divided by the number of non-graduates.

The general method is to identify the conditioning event, restrict the counting to it, and then divide the favourable count by the size of the restricted group.

Complementary event probability

The complement of an event A is \overline{A} with P(\overline{A}) = 1 - P(A). This is useful whenever the 'at least' or 'not' version is harder to count than its opposite. If the event is estimated by relative frequency, for example a batsman hitting a six in a number of balls, then the probability of not hitting a six is one minus the observed frequency.

Complements are also used in counting. Among the numbers 1 to 50, the composites are found by subtracting the primes and the number 1 from the total, because 1 is neither prime nor composite.

For two events, the four cells A\cap B, A\cap\overline{B}, \overline{A}\cap B, \overline{A}\cap\overline{B} partition the sample space. The three cells other than A\cap B therefore have probabilities that sum to 1 - P(A\cap B). A similar complement argument handles 'at most' conditions: the probability of at most four tails in five tosses is 1 minus the probability of five tails.

For a set operation, the complement of \overline{A}\cup B is A\cap\overline{B}, so P(\overline{A}\cup B) = 1 - P(A\cap\overline{B}), and P(A\cap\overline{B}) = P(A) - P(A\cap B).

Probability with number formation

When digits are arranged without repetition, the number of arrangements of n distinct digits taken all together is n!, and each arrangement is equally likely when a number is chosen at random. Divisibility conditions are then checked on the digits. A number is divisible by 3 exactly when its digit sum is divisible by 3, and this is the key test for a set of formed numbers.

When two digits are chosen from a set and multiplied, the choice is unordered, so the sample space has \binom{n}{2} equally likely pairs. A product ends in 0 only when it contains both factors 2 and 5. Counting the pairs that supply both factors gives the favourable count.

For a pair of distinct numbers chosen from a set, the sum is divisible by a given integer when the two numbers fall in complementary residue classes. Counting the pairs whose sum is 4, 8, 12 or 16 and adding them gives the favourable count over \binom{10}{2}.

An inequality involving a number and its reciprocal can also decide which numbers are favourable. For x > 0, x + \frac{1}{x} - 2 = \frac{(x-1)^2}{x} \geq 0, with equality only at x = 1. Every number except 1 therefore satisfies the strict inequality, and the probability is the count of such numbers divided by the total.

Probability of picking numbers at random

When numbers are picked at random, every selection is equally likely, so the probability of an event is the number of favourable selections divided by the total number of selections. If two distinct numbers are chosen from n numbers and order is ignored, the total is \binom{n}{2}. A pair is unordered, so \{3,7\} is counted once, not twice. For products, only the units digits matter, because the units digit of ab depends only on the units digits of a and b. Among distinct numbers from 1 to 9, a product ends in 1 only when the units digits multiply to a number ending in 1, and the repeated pair 9\times 9 is not allowed. In 'product of two equals the third' problems, the number 1 must be excluded as a factor, since 1\times b=b repeats b. Favourable triples have the form \{a,b,ab\} with 2\le a<b and ab\le n, and their count is divided by \binom{n}{3}. The number of multiples of k among 1 to N is \lfloor N/k\rfloor. For 'divisible by a or b', use n(A\cup B)=n(A)+n(B)-n(A\cap B), where n(A\cap B) counts the multiples of the lcm. The 'neither' count is the total minus this union. Parity problems use the same logic. In a block of consecutive integers with equal odd and even counts, a pair has odd sum exactly when one number is odd and the other is even, so the favourable count is (number of odd)\times(number of even). Example: from 1 to 6 there are 3 odd and 3 even numbers, so P(\text{sum odd})=\frac{3\times 3}{\binom{6}{2}}=\frac{9}{15}=\frac{3}{5}.

Independent and mutually exclusive events

Two events A and B are mutually exclusive when they cannot occur together, so P(A\cap B)=0 and P(A\cup B)=P(A)+P(B). They are independent when one does not affect the other, which is tested by P(A\cap B)=P(A)P(B). Since P(A\cap B)=0 for exclusive events, two events with positive probabilities cannot be both exclusive and independent. To test independence from given data, first find the missing probability using P(A\cup B)=P(A)+P(B)-P(A\cap B), then compare P(A)P(B) with P(A\cap B). For independent events X and Y with P(X)=p and P(Y)=q, the basic combinations are: both occur pq; neither occurs (1-p)(1-q); exactly one occurs p(1-q)+q(1-p)=p+q-2pq; at least one occurs 1-(1-p)(1-q)=p+q-pq. A common error is to use p+q-pq for 'exactly one', which is the 'at least one' value. For several independent events, the probability that all occur is the product of their probabilities. Thus three independent events each with probability 0.6 all occur with probability (0.6)^3=0.216. Example: if P(X)=0.5 and P(Y)=0.2, then exactly one occurs with probability 0.5(0.8)+0.2(0.5)=0.5.