# Probability question: marbles in a jar

A jar contains $3$ yellow marbles, $4$ red marbles, $10$ green marbles. and $4$ blue marbles. What is the probability that the first marble picked at random is blue and that the second marble is green and that the third marble picked is yellow, assuming that the marbles are put back into the jar after every time they are picked?

My attempt:

Probability the first marble is blue: $\frac{4}{21}$.

Probability the second marble is green: $\frac{10}{20} = \frac{1}{2}$

Probability the third marble is yellow: $\frac{3}{19}$

I don't think this is right though. Can someone help me please? Thank you.

• I think one of the key points of the question is "marbles are put back after every time they are picked"... – abiessu Jan 12 '15 at 3:01
• All the denominators are to be $21$ and then multiply the fractions together to get the resulting probability since each draw is independent. – user60887 Jan 12 '15 at 3:12
• Oh so it would be $(4/21) * (10/21) * (3/21)$? – NewtoProb Jan 12 '15 at 3:14
• @NewtoProb Yes. – turkeyhundt Jan 12 '15 at 3:14

From independence (due to replacement of drawn marbles) we obtain the answer as $$\frac{4}{21}\times \frac{10}{21}\times \frac{3}{21} = \frac{120}{9261}.$$