Classical probability

When all outcomes in a sample space are equally likely , and there are finitely many , we can compute probability simply by counting: $\boxed{P(A) = \frac{|A|}{|S|} = \frac{\text{number of outcomes in } A}{\text{number of outcomes in } S}.}$

This is the classical (or Laplacian) definition of probability. It works for coin tosses, die rolls, card draws, and most basic problems.

When does this apply?

The classical formula requires:

1. Finite sample space. $|S| < \infty$.
2. Equally likely outcomes. Symmetry usually justifies this , a fair coin, a balanced die, well-shuffled cards.

If outcomes are not equally likely (a biased coin), or the sample space is infinite, you need other tools.

Counting tools

Class XI Chapter 6 gave you permutations and combinations , exactly the tools for counting.

• Number of ways to choose $r$ from $n$ without order: $\binom{n}{r}$.
• Number of ways to arrange $r$ items from $n$: $P(n, r) = n!/(n-r)!$.
• Multiplication principle: if there are $m$ ways to do A and $n$ ways to do B (independently), then $m \cdot n$ ways to do both.

Use these to count $|A|$ and $|S|$.

Worked examples

Example 1. A fair die is rolled. Probability of getting an even number?

$S = \{1, 2, 3, 4, 5, 6\}$, $|S| = 6$. $A = \{2, 4, 6\}$, $|A| = 3$. $P(A) = 3/6 = 1/2$.

Example 2. Two dice are rolled. Probability of a sum of $7$?

$|S| = 36$. $A =$ sum $= 7$ $= \{(1,6), (2,5), \dots, (6,1)\}$. $|A| = 6$. $P(A) = 6/36 = 1/6$.

Example 3. Two cards drawn from a deck. Probability both are kings.

$|S| = \binom{52}{2} = 1326$. $|A| = \binom{4}{2} = 6$. $P(A) = 6/1326 = 1/221$.

Example 4. A bag has $5$ red and $4$ blue balls. Two balls drawn without replacement. Probability both red?

$|S| = \binom{9}{2} = 36$. $|A| = \binom{5}{2} = 10$. $P(A) = 10/36 = 5/18$.

Example 5. Four people sit at random in a row of $4$ chairs. Probability the two specific people $X, Y$ sit together?

$|S| = 4! = 24$. For $|A|$: treat $X, Y$ as a single unit; the unit and the other $2$ people in $3! = 6$ arrangements; $X, Y$ themselves in $2$ orders. So $|A| = 6 \cdot 2 = 12$. $P(A) = 12/24 = 1/2$.

Try it yourself

1. Toss a coin. Probability of heads?
2. Roll a die. Probability of a prime number?
3. Roll two dice. Probability of doubles?
4. Draw one card. Probability of a queen?
5. Draw one card. Probability of a red face card?
6. Three coins tossed. Probability of exactly two heads?
7. Two dice. Probability sum is divisible by $3$?
8. From a deck, $5$ cards drawn. Probability all four aces are among them.
9. From $\{1, 2, \dots, 100\}$ a number is chosen. Probability it is divisible by $5$ or $7$.
10. A box has $4$ defective and $6$ good bulbs. $3$ drawn. Probability all three good?
11. Three dice rolled. Probability all show the same number?
12. From a deck, $13$ cards are drawn. Probability they are all of one suit (hint: $4$ suits $\times$ $\binom{13}{13} = 4$ favourable; $|S| = \binom{52}{13}$).

Pitfalls / Tricks

• Always verify equally likely.
• Be careful about "ordered" vs "unordered" , they give different $|S|$ but should give the same probability if used consistently.
• Use complement: sometimes $P(A^c)$ is easier than $P(A)$.
• Insight. Classical probability is counting in disguise. Master Chapter 6 (permutations and combinations) and most of these problems are routine.