A bag of marbles contains only green and red marbles. All of the marbles are the same size. If the probability of selecting a green marble is 1/2. Which is most likely to be the number of green and red marbles in the bag?
A.2 green 4 red
B.4 green 4 red
C.2 green 1 red
D.1 green 3 red
E. 3 green 6 red
I don't know which one
A and E both equal 1/2
Oops - you're not understanding the problem.
If there are two marbles, one green and one red, your chance of drawing a green marble is 1 out of 2 or 1/2.
You have to have the same number of each color of marbles to have the odds be 1/2.
Right.
THank you
You're welcome.
To determine which option is most likely, we need to compare the probabilities of selecting a green marble for each option.
Let's calculate the probabilities for each option:
A. 2 green 4 red:
Probability of selecting a green marble = 2/(2+4) = 2/6 = 1/3
B. 4 green 4 red:
Probability of selecting a green marble = 4/(4+4) = 4/8 = 1/2
C. 2 green 1 red:
Probability of selecting a green marble = 2/(2+1) = 2/3
D. 1 green 3 red:
Probability of selecting a green marble = 1/(1+3) = 1/4
E. 3 green 6 red:
Probability of selecting a green marble = 3/(3+6) = 3/9 = 1/3
Comparing the probabilities, we see that the options with the highest probability of selecting a green marble are B (1/2) and E (1/3).
Therefore, the most likely scenario for the number of green and red marbles in the bag is option B: 4 green marbles and 4 red marbles.