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.
a. P(BG)=
b. P(BB)=
c. P(GG)=
d. P(GB)=
I am confused on the concept.
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
b. P(BB)=4/7*3/7=12/42=2/7
c.P(GG)=3/7*2/7=6/42=1/7
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)
To find the probabilities of the various outcomes, we can use the concept of conditional probability. In this case, we are drawing candies from the box without replacement, meaning that the number of candies in the box decreases each time.
Let's break down the question to find the probabilities:
a. P(BG) represents the probability of selecting a blue candy first and a green candy second.
To find P(BG), we need to consider the probability of selecting a blue candy first, which is (3 blue candies) / (7 total candies) = 3/7.
After selecting a blue candy, the number of remaining candies becomes 6 (since we didn't replace the initial candy). Now, the probability of selecting a green candy is (4 green candies) / (6 remaining candies) = 4/6.
To find the overall probability of selecting a blue candy first and a green candy second, we multiply these two probabilities together: P(BG) = (3/7) * (4/6) = 12/42.
Simplifying 12/42 gives P(BG) = 2/7.
b. P(BB) represents the probability of selecting two blue candies consecutively.
The probability of selecting the first blue candy is the same as before, 3/7.
Now, the number of blue candies remaining is 2, while the total remaining candies become 6.
The probability of selecting a second blue candy, without replacement, is therefore (2 blue candies) / (6 remaining candies) = 2/6.
To find the overall probability of selecting two blue candies consecutively, we multiply these two probabilities together: P(BB) = (3/7) * (2/6) = 6/42.
Simplifying 6/42 gives P(BB) = 1/7.
c. P(GG) represents the probability of selecting two green candies consecutively.
The probability of selecting the first green candy is (4 green candies) / (7 total candies) = 4/7.
After selecting a green candy, the remaining number of candies is 6, and the remaining number of green candies is 3 - 1 = 2.
The probability of selecting a second green candy, without replacement, is (2 green candies) / (6 remaining candies) = 2/6.
To find the overall probability of selecting two green candies consecutively, we multiply these two probabilities together: P(GG) = (4/7) * (2/6) = 8/42.
Simplifying 8/42 gives P(GG) = 4/21.
d. P(GB) represents the probability of selecting a green candy first and a blue candy second.
The probability of selecting the first green candy is the same as before, 4/7.
After selecting a green candy, the remaining number of candies is 6, and the number of remaining blue candies is 3.
The probability of selecting a blue candy, without replacement, is (3 blue candies) / (6 remaining candies) = 3/6.
To find the overall probability of selecting a green candy first and a blue candy second, we multiply these two probabilities together: P(GB) = (4/7) * (3/6) = 12/42.
Simplifying 12/42 gives P(GB) = 2/7.
In summary, the probabilities are as follows:
a. P(BG) = 2/7
b. P(BB) = 1/7
c. P(GG) = 4/21
d. P(GB) = 2/7