Question

All that is left in a packet of candy are 6 reds, 3 greens, and 3 blues.

(a)What is the probability that a random drawing yields a blue followed by a green assuming that the first candy drawn is put back into the packet?

(b)Are the events 'blue' and 'green' independent?

Answer 1

a) There is a 3/12 chance of getting a blue the first time and a 3/12 chance of getting a green the second time. Multiplying these

gives (3/12)(3/12) =(1/4)(1/4) = 0.0625

since the first candy drawn is put back into the packet (with replacement)

b) P(getting a blue) = 3/12 = 0.25 and P(getting a green) = 3/12 = 0.25

since P(getting a blue) * P(getting green) = 0.25*0.25 = 0.0625 = P(getting a blue and getting green)

Therefore "blue" an "green" are independent