ironfern @ docs ~/math/grade-9/combinatorics/classical-probability $

Classical Definition of Probability

To find the probability of certain events, it is not always necessary to conduct experiments or observations. Sometimes it suffices to rely on life experience and common sense.

Certain and Impossible Events

📐Definition — Certain Event

An event that is guaranteed to occur under the given set of conditions in every trial is called a certain (or sure) event. The probability of such an event is 1: if AA is a certain event, then P(A)=1P(A) = 1.

📐Definition — Impossible Event

An event that cannot occur under the given set of conditions in any trial is called an impossible event. The probability of such an event is 0: if AA is an impossible event, then P(A)=0P(A) = 0.

Example 1 — Balls in a Box

Statement. A box contains 10 red balls. What is the probability that a randomly drawn ball is red? yellow?

Solution. The probability of drawing a red ball is 1 (certain event). Since there are no yellow balls in the box, drawing a yellow ball is impossible — P(A)=0P(A) = 0.

Equally Likely Outcomes and the Classical Definition

Example 2 — Tossing a Coin

Statement. A fair coin is tossed once. What is the probability of getting heads?

Solution. This experiment can produce only one of two outcomes: heads or tails. Neither outcome is favored over the other. Such outcomes are called equally likely, and the corresponding random events are called equiprobable. It is therefore natural to consider the probability of each event — “heads” and “tails” — equal to 12\dfrac{1}{2}.

Example 3 — Rolling a Die

Statement. A die is rolled once. What is the probability of rolling a 4?

Solution. This experiment can yield one of six outcomes: 1, 2, 3, 4, 5, or 6 dots. All outcomes are equally likely. Therefore the probability of rolling a 4 is 16\dfrac{1}{6}.

In each of the examples above, the probability of event AA can be calculated using the formula:

P(A)=mnP(A) = \frac{m}{n}
📐Definition — Classical Definition of Probability

If an experiment can result in one of nn equally likely outcomes, of which mm are favorable to event AA, then the probability of event AA is defined as the ratio mn\dfrac{m}{n}.

P(A)=mnP(A) = \frac{m}{n}

This definition of probability is called classical.

Note — Limitations of the Classical Definition

Note that when the conditions of an experiment are such that its outcomes are not equally likely, the classical definition of probability cannot be applied.

Properties of Probability

From the definition it follows that:

0P(A)1.0 \leqslant P(A) \leqslant 1.
  • If AA is an impossible event, then m=0m = 0 and P(A)=0P(A) = 0.
  • If AA is a certain event, then m=nm = n and P(A)=1P(A) = 1.
Example 4 — Lottery Ticket

Statement. Suppose 100,000 lottery tickets have been issued, 20 of which are winners. What is the probability of winning when buying one ticket?

Solution. Out of 100,000 equally likely outcomes, 20 lead to a win. Therefore the probability of winning is

20100000=15000.\frac{20}{100\,000} = \frac{1}{5000}.

Answer: 15000\dfrac{1}{5000}.

Example 5 — Two Dice

Statement. Two dice — one blue and one yellow — are rolled simultaneously. What is the probability of rolling two sixes?

Solution. This experiment can produce 36 equally likely outcomes, of which only one is favorable (both dice showing 6).

Therefore the desired probability is 136\dfrac{1}{36}.

Answer: 136\dfrac{1}{36}.

Example 6 — D'Alembert's Problem — Tossing Two Coins

Statement. Two identical coins are tossed simultaneously. What is the probability that at least one shows heads?

Solution. To make all outcomes equally likely, we distinguish the coins by numbering them. Then there are four equally likely outcomes. In the first three outcomes, at least one coin shows heads.

Therefore the probability is 34\dfrac{3}{4}.

Answer: 34\dfrac{3}{4}.

Example 7 — Families with Two Children

Statement. Consider all families with two children in which at least one child is a boy. What is the probability that a randomly chosen such family has two boys?

Solution. The set of conditions in our experiment gives three equally likely outcomes:

  • older child — boy, younger child — boy;
  • older child — boy, younger child — girl;
  • older child — girl, younger child — boy.

Therefore the desired probability is 13\dfrac{1}{3}.

Answer: 13\dfrac{1}{3}.

Exercise — Classical Probability Problems
  1. A basket contains 10 red and 15 green apples. What is the probability of randomly picking a pear? an apple?

  2. What is the probability that when rolling a die, the number of dots equals: 1) one; 2) three; 3) an odd number?

  3. One number is chosen at random from the natural numbers 1 through 30. What is the probability that it is: 1) prime; 2) a divisor of 18; 3) a perfect square?