Problem 124

Problem 124

A box contains two black balls and three gold balls. Two balls are randomly drawn in succession from the box.

a. If there is no replacement, what is the probability that both balls are black?

b. If there is replacement before the second draw, what

is the probability that both balls are black?

Solution

Total balls in the bag are 2+3=5

Total ways to choose two balls =⁵c₂=10

Ways to choose one black ball = ²c₁=2

P(one black ball brawn) = $\frac{2}{10} = \frac{1}{5}$

Ways to choose other black ball is =1

And total ways to choose one ball between remaining 9 balls = ⁹c₁=9

So P(another black ball) = $\frac{1}{9}$

So final probability = $\frac{1}{5} \ast \frac{1}{9} = \frac{1}{45}$

Total balls in the bag are 2+3=5

Total ways to choose two balls =⁵c₂=10

Total favorable cases (replacement)=(²c₁+²c₁)=4

P(both black )= $\frac{4}{10} = \frac{2}{5}$