First slide

Introduction to probability

Question

A bag contains b blue balls and r red balls. If two balls are drawn at random, the probability of drawing two red balls is five times the probability of drawing two blue balls. Furthermore, the probability of drawing one ball of each color is six times the probability of drawing two blue balls. Then

Moderate

A

b + r = 9
B

br = 18
C

$| b - r | = 4$
D

b/r = 2

Solution

Let the number of red and blue balls be r and b, respectively Then, the probability of drawing two red balls is
$p_{1} = \frac{^{r} C_{2}}{{}^{r + b}C_{2}} = \frac{r (r - 1)}{(r + b) (r + b - 1)}$
The probability of drawing two blue balls is
$p_{2} = \frac{^{b} C_{2}}{^{r + b} C_{2}} = \frac{b (b - 1)}{(r + b) (r + b - 1)}$
The probability of drawing one red and one blue ball is
$p_{3} = \frac{^{r} C_{1} \times^{b} C_{1}}{{}^{r + b}C_{2}} = \frac{2 br}{(r + b) (r + b - 1)}$
By hypothesis, p₁= 5p₂ and p₃ = 6p₂.Therefore,
$r (r - 1) = 5 b (b - 1) and 2 br = 6 b (b - 1)$
$\Rightarrow r = 6, b = 3$

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