A box contains twelve candies, five reds, four greens, and three blues. You select one piece at random, then replace it, then you select another. If
P(BG) is the probability of selecting a blue first and a green second, find the following probabilities.
a.
P(RG)=
b. P(BB)=
c. P(GB)=
d. P(not RR)=
I am confused do not know how to start.
Did you not look at and follow the solution and explanations I gave you here?
https://www.jiskha.com/questions/1775289/A-box-contains-twelve-candies-five-reds-four-greens-and-three-blues-You-select
https://www.jiskha.com/questions/1775288/A-box-contains-seven-candies-four-of-which-are-green-and-three-of-which-are-blue
To solve these probability problems, we need to understand the concept of probability. Probability measures the likelihood of an event occurring. In this case, we have a box with twelve candies, and we are interested in finding the probability of certain color combinations.
Before we start calculating probabilities, it's important to note that when we replace the candy back into the box after selecting it, the total number of candies remains the same for each selection.
Now, let's solve each problem step by step:
a. P(RG): This means the probability of selecting a red candy first and a green candy second.
To calculate this probability, we need to find two separate probabilities and multiply them together. The probability of selecting a red candy first is P(R) = 5 red candies / 12 total candies = 5/12. The probability of selecting a green candy second is P(G) = 4 green candies / 12 total candies = 4/12.
So, P(RG) = P(R) * P(G) = (5/12) * (4/12) = 20/144 = 5/36.
b. P(BB): This means the probability of selecting two blue candies in a row.
Since we're replacing each candy back into the box after selecting, the probability of selecting a blue candy on each draw remains the same. Thus, the probability of selecting a blue candy is P(B) = 3 blue candies / 12 total candies = 3/12 = 1/4.
To calculate P(BB), we multiply the probability of selecting a blue candy twice: P(BB) = P(B) * P(B) = (1/4) * (1/4) = 1/16.
c. P(GB): This means the probability of selecting a green candy first and a blue candy second.
Again, we find the probabilities of each event separately. The probability of selecting a green candy first is P(G) = 4 green candies / 12 total candies = 4/12 = 1/3. The probability of selecting a blue candy second is P(B) = 3 blue candies / 12 total candies = 3/12 = 1/4.
So, P(GB) = P(G) * P(B) = (1/3) * (1/4) = 1/12.
d. P(not RR): This means the probability of not selecting a red candy on either draw.
To calculate this probability, we need to find the complement of the probability of selecting a red candy on either draw. The probability of selecting a red candy on the first draw is P(R) = 5 red candies / 12 total candies = 5/12. Since we replace the candy back into the box, the probability of not selecting a red candy on the first draw is P(not R) = 1 - P(R) = 1 - 5/12 = 7/12.
Using the same reasoning, the probability of not selecting a red candy on the second draw is also P(not R) = 7/12.
To find P(not RR), we multiply the probabilities of not selecting a red candy on each draw: P(not RR) = P(not R) * P(not R) = (7/12) * (7/12) = 49/144.
So, the probabilities are:
a. P(RG) = 5/36
b. P(BB) = 1/16
c. P(GB) = 1/12
d. P(not RR) = 49/144