# The probability of select marbles from two jars?

I don't know how to get the right answer. I know that the probability of selecting a red marble from the mayonnaise jar is 6/10. There are two possibilities of selecting a red marble from the jelly jar. If the first selection is red, the probability is 3/8, otherwise, the probability is 2/8.

A mayonnaise jar contains 6 red marbles and 4 blue marbles. A jelly jar contains 2 red marbles and 5 blue marbles. One marble is randomly selected from the mayonnaise jar and placed in the jelly jar. A marble is then randomly selected from the jelly jar. What is the probability that the selected marble is red? Express your answer as a common fraction.

• Ok, we know that the probability of getting a red marble from the jelly jar is 2/5 BEFORE we add another marble to the jelly jar. What do we have to add or subtract from 2/5? – wesssg Mar 1 '17 at 5:15

You can split it up into two cases. First if the marble transferred from the mayo jar is red (call this event $R_m$) then conditional on that, the probability that the marble selected from the jelly jar is red is $$P(R_j|R_m) = \frac{3}{8}$$ since there are $3$ red marbles and five blue ones. If the marble transferred is blue instead, we have $$P(R_j|B_m) = \frac{2}{8}$$ since then there are two red and six blue.

Now we know that the probability that a red is transferred is $P(R_m) = 6/10$ since there are six red and four blue in the mayo jar. So the law of total probability gives $$P(R_j) = P(R_j|R_m)P(R_m) + P(R_j|B_m)P(B_m) = \frac{3}{8}\frac{6}{10} + \frac{2}{8}\frac{4}{10} = \frac{13}{40}.$$

• You shouldn't really do an entire homework problem. Just show the way and leave something for the student to do on their own. – Graham Kemp Mar 1 '17 at 5:46