A box contains 10 black and 10 white balls. What is the probability of drawing 2 balls of the same colour?

### Concept & Logic The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. When drawing items without replacement, we use combinations. The probability of drawing 2 balls of the same color means drawing either 2 black balls OR 2 white balls. $$ P(\text{same color}) = P(\text{2 black}) + P(\text{2 white}) $$ ### Step-by-Step Solution * **Total balls:** 10 black + 10 white = 20 balls. * **Total ways to draw 2 balls:** ${}^{20}C_2 = \frac{20 \times 19}{2 \times 1} = 190$. * **Ways to draw 2 black balls:** ${}^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$. * **Ways to draw 2 white balls:** ${}^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$. * **Total favorable ways:** 45 + 45 = 90. * **Probability:** $\frac{90}{190} = \frac{9}{19}$. ### Exam Strategy & Shortcut For finding the probability of 2 items of the same color from $n_1$ and $n_2$ items, use the formula $\frac{{n_1}C_2 + {n_2}C_2}{{N}C_2}$. In this case, since the number of black and white balls are equal (10 each), it's simply $2 \times \frac{{}^{10}C_2}{{}^{20}C_2} = 2 \times \frac{45}{190} = \frac{9}{19}$. ### Common Pitfall A common mistake is to calculate the probability with replacement (i.e., $\frac{10}{20} \times \frac{10}{20} + \dots$). Unless "with replacement" is specified, drawing multiple items at once implies "without replacement". ### Final Answer Therefore, the correct answer is **$\frac{9}{19}$**.