# 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

## 2 Answers

You are decreasing the denominators of your fractions, but you shouldn't do that because you are putting the marbles back after each pick, so there are always 21 marbles in the jar.

• I noticed that but maybe I misunderstood the question. I thought after putting the marbles back I draw out 2 marbles and then put them back and draw out 3 marbles respectively. – NewtoProb Jan 12 '15 at 3:12
• Normally problems worded this way mean that you pick a marble, note it, drop it back in, go to the next trial. – turkeyhundt Jan 12 '15 at 3:13
• I see that makes sense. Thanks – NewtoProb Jan 12 '15 at 3:14

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