A box contains seven candies, four of which are green, and three of which are blue. You select one piece at random, then without replacing the first one, you select another. If P(BG) is the probability of selecting a blue first and a green second, find the following probabilities.

Question

a. P(BG)=

b. P(BB)=

c. P(GG)=

d. P(GB)=

I am confused on the concept.

Answer 1

Not sure at the depth of study you are for this topic, but here is a simple way:

to get a green:
out of the 7 candies, 4 of them are green, so the prog(a green on first draw)
= 4/7
so now there are 6 candies left to pick from of which 3 are blue, so the prob(a blue for 2nd draw) = (3/6)

So the prob(first a green, then a blue)
= (4/7)(1/2) = 4/14 = 2/7

let me know how you worked the others

Answer 2

b. P(BB)=4/7*3/7=12/42=2/7
c.P(GG)=3/7*2/7=6/42=1/7

Answer 3

No, those answers would be correct if you replaced the candies.

Did you not look at my solution and see how the concept of replacement/non-replacement must be handled?

I noticed in b) you have P(BB)=4/7*3/7=12/42=2/7
your first calculation is 4/7*3/7, which is wrong, but then your answer to this
is 12/42, which is the correct answer but the whole question, but is not derived from your calculation.
If I were marking this on a test, I would deduct marks twice,

You have the same situation in c)