Question

There are 10 black and 5 white balls in a bag. Two balls are taken out, one after another, and the first ball is not placed back before the second is taken out. Assume that the drawing of each ball from the bag is equally likely. What is the probability that both the balls drawn are black? 

Hint:

For drawing without replacement, multiply the probabilities of each event occurring one after the other.

Answer 1

Step 1: The total number of balls is \( 10 + 5 = 15 \). The probability of drawing a black ball on the first draw is: \[ P(\text{first ball black}) = \frac{10}{15}. \] Step 2: After the first ball is drawn, the total number of balls left is 14. The probability of drawing a second black ball is: \[ P(\text{second ball black}) = \frac{9}{14}. \] Step 3: The probability that both balls are black is the product of these two probabilities: \[ P(\text{both black}) = \frac{10}{15} \times \frac{9}{14} = \frac{90}{210} = \frac{3}{7}. \]