SOLUTION: A bag contains 8 red marbles, 4 white marbles, and 5 blue marbles. Find P(red and blue). (Don't just tell me the answer, show me your work!)

Question 1010750: A bag contains 8 red marbles, 4 white marbles, and 5 blue marbles. Find P(red and blue). (Don't just tell me the answer, show me your work!)
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
A bag contains 8 red marbles, 4 white marbles, and 5 blue marbles. Find P(red and blue).

Solution:
We will assume that only two marbles are drawn from the bag, and hence there are two cases:
A. The first marble is returned in the bag before drawing the second.
B. The first marble is not returned in the bag before drawing the second.
Let
R=event of drawing a red
B=event of drawing a blue.
Therefore the successful events are RB or BR.

A. First marble returned to the bag (with replacement).
In this case, marbles are drawn with 8+4+5=17 in the bag.
P(RB)=(8/17)*(5/17)=40/289 (multiplied because events are independent)
P(BR)=(5/17)*(8/17)=40/289
Probability of drawing BR or RB is
P(RB or BR)=40/289+40/289=80/289
(probabilities are added because RB and BR are mmutually exclusive)

B. First Marble not returned (without replacement).
Here, for the first draw, there are 17 marbles in the bag, and only 16 marbles left in the second draw.
P(RB)=(8/17)*(5/16)=40/272 (multiplied because events are independent)
P(BR)=(5/17)*(8/16)=40/272
Probability of drawing BR or RB is
P(RB or BR)=40/272+40/289=80/272
(probabilities are added because RB and BR are mutually exclusive)

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