Question

A bag contains 6 red, 4 blue and 8 green marbles. How many marbles of each color should be added so that the total number of marbles is 27, but the probability of randomly selecting one marble of each color remains unchanged.

Probability
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asked 2020-11-26
A bag contains 6 red, 4 blue and 8 green marbles. How many marbles of each color should be added so that the total number of marbles is 27, but the probability of randomly selecting one marble of each color remains unchanged.

Answers (1)

2020-11-27
The bag contains 6 red, 4 blue and 8 green marbles, which er 6+4+8=18 marbles in total.
The probability is the number of favorable outcomes divided by the number of possible outcomes:
P(red)=(# of favorable outcomes)/(# of favorable outcomes)=\(\displaystyle\frac{{6}}{{18}}=\frac{{1}}{{3}}\)
P(blue)=(# of favorable outcomes)/(# of favorable outcomes)=\(\displaystyle\frac{{4}}{{18}}=\frac{{2}}{{9}}\)
P(green)=(# of favorable outcomes)/(# of favorable outcomes)=\(\displaystyle\frac{{8}}{{18}}=\frac{{4}}{{9}}\)
We now want to have a bag with 27 marbles instead (while the probabilities remain the same) and thus the number of marbles of each color can the be obtained by multiplying the total number of marbles by the probability of that color.
Red:\(\displaystyle{27}\cdot{\left(\frac{{1}}{{3}}\right)}={9}\)
Blue:\(\displaystyle{27}\cdot{\left(\frac{{2}}{{9}}\right)}={3}\cdot{2}={6}\)
Green:\(\displaystyle{27}\cdot{\left(\frac{{4}}{{9}}\right)}={3}\cdot{4}={12}\)
Thus the bag should contain 9 red marbles, 6 blue marbles and 12 green marbles.
Since the bag originally contained 6 red marbles, 4 blue marbles and 8 green marbles, we need to add 3 red marbles, 2 blue marbles and 4 green marbles.
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