# Probability question at least one without replacement.

Title says it all, it’s a bit confusing to understand.

A bag contains 13 marbles 6 red, 3 yellow, and 4 green. If 3 marbles are chosen at random (without replacement) then what is the probability that at least one of them is yellow.

This is what I got. 1-(3/13)(2/12)(1/11).

I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11

• math.stackexchange.com/questions/697433/… – lab bhattacharjee Dec 4 '18 at 16:19
• What have you tried so far? – user3482749 Dec 4 '18 at 16:39
• This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct – J.sant Dec 4 '18 at 16:41
• I was asking for a thought process, not a number. – user3482749 Dec 4 '18 at 16:51
• Well, I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11 – J.sant Dec 4 '18 at 16:58

HINT

1. Find the probability that none are yellow.
2. Remember that if $$A$$ is any event, then $$\mathbb{P}[A] = 1-\mathbb{P}[\bar{A}]$$

UPDATE

1. The first time you grab a yellow ball with a chance of 3/13. You want the negation of this, so the chance of that is $$1-3/13 = 10/30$$.
2. The second time, you grab a yellow (assuming first time you did not) with a chance 3/12. Third time it will be 3/11.
3. You want to draw yellow in neither of the tries, the probability of that is $$\left[1 - \frac{3}{13}\right] \times \left[1 - \frac{3}{12}\right] \times \left[1 - \frac{3}{11}\right] = \frac{10 \cdot 9 \cdot 8}{13 \cdot 12 \cdot 11} = \frac{60}{143}$$
4. Finally you want the probability that any draw results in a yellow, which is $$1 - \frac{60}{143} = \frac{83}{143} \approx 58\%.$$
• This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct? – J.sant Dec 4 '18 at 16:35
• @J.sant No, you have to calculate $1-P("\color{blue}{\textrm{No yellow}} \ \textrm{marble is chosen}")$. The probability is approximately $58\%$ – callculus Dec 4 '18 at 17:35
• So 1-10/13. That’s the probability of none yellow. Is that correct? – J.sant Dec 4 '18 at 18:22
• @J.sant Not really. You chose your first non-yellow marble: $\frac{10}{13}$. Then you choose your second non-yellow marble: $\frac{9}{12}$. And lastly you choose your third non-yellow marble: $\frac{8}{11}$. Now subtract the product from $1$. – callculus Dec 4 '18 at 18:53
• I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1?? – J.sant Dec 4 '18 at 20:01