Question

All that is left in a packet of candy are 5 reds, 2 greens, and 2 blues. (a)What is the probability that a random drawing yields a green followed by a blue assuming that the first candy drawn is put back into the packet?

Answer 1

Answer:

4/81

Step-by-step explanation:

The candies in the packet are:

r = 5 (number of red candies)

g = 2 (number of green candies)

b = 2 (number of blue candies)

The total number of candies in the packet at the beginning is:

n = r + g + b = 5 + 2 +2 = 9

Therefore, at the first attempt, the probability of drawing a green candy is:

$p(g)=\frac{g}{n}=\frac{2}{9}$

Then, the first candy is placed back in the packet, so still

n = 9

Therefore, at the second attempt, the probability of drawing a blue candy is

$p(b)=\frac{b}{n}=\frac{2}{9}$

Therefore, the probability that a random drawing yields a green followed by a blue is:

$p(gb)=p(g)p(b)=\frac{2}{9}\cdot \frac{2}{9}=\frac{4}{81}$