A box contains ten balls: three white and seven black. One…

A box contains ten balls: three white and seven black. One ball is drawn at random from the box. Determine the probability that the drawn ball will be: $1^{\circ}$ white; $2^{\circ}$ black.

Step 1: The classical definition of probability

If an experiment has $n$ equally likely outcomes and exactly $m$ of them lead to the event $A$, then

\[ P(A) = \frac{m}{n} = \frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}. \]

Step 2: The outcomes

Imagine the balls are numbered $1,\dots,10$, with balls $1$–$3$ white and $4$–$10$ black. The outcome of the experiment is the number of the drawn ball:

\[ \Omega=\{1,2,\dots,10\}, \qquad n = 10. \]

All ten outcomes are equally likely because the ball is drawn at random (the balls are identical in size and well mixed).

Step 3: $1^{\circ}$ — a white ball

Favourable outcomes: $\{1,2,3\}$, so $m=3$.

\[ \boxed{P(\text{white}) = \frac{3}{10} = 0.3.} \]

Step 4: $2^{\circ}$ — a black ball

Favourable outcomes: $\{4,\dots,10\}$, so $m=7$.

\[ \boxed{P(\text{black}) = \frac{7}{10} = 0.7.} \]

Step 5: Check

"White" and "black" are opposite events (every ball is one or the other, never both), so their probabilities must add up to $1$:

\[ 0.3 + 0.7 = 1. \ \checkmark \]

We could therefore also compute $P(\text{black}) = 1 - P(\text{white})$.

Step 6: Meaning

If the experiment is repeated many times (drawing a ball, noting its colour and putting it back), a white ball will appear in about $30\%$ of the draws and a black ball in about $70\%$.