Question

1. jaylan has a fun-size bag of sweettarts™. inside the bag are 4 pink, 1 blue, 2 yellow, 3 purple and 2 green sweettarts™. what is the probability of drawing a yellow sweettart three times in a row if each sweettart™ selected is eaten before the next sweettart™ is drawn?

Explanation:

Step1: Calculate total number of SweetTarts

First, we sum up the number of each color. The number of pink is 4, blue is 1, yellow is 2, purple is 3, and green is 2. So total \( n = 4 + 1 + 2 + 3 + 2 = 12 \).

Step2: Probability of first yellow

The number of yellow is 2, so the probability of drawing a yellow first is \( P_1=\frac{2}{12}=\frac{1}{6} \).

Step3: Probability of second yellow (after eating first)

After eating one yellow, the number of yellow left is \( 2 - 1 = 1 \), and total number left is \( 12 - 1 = 11 \). So probability \( P_2=\frac{1}{11} \).

Step4: Probability of third yellow (after eating second)

After eating two yellows, number of yellow left is \( 1 - 1 = 0 \)? Wait, no, wait. Wait, initial yellow is 2. Wait, first draw: 2 yellow out of 12. Second draw: after eating one yellow, 1 yellow out of 11. Third draw: after eating two yellows, 0 yellow out of 10? Wait, no, the problem says "drawing a yellow SweetTart three times in a row". But there are only 2 yellow ones. Wait, maybe I misread. Wait, the problem: "drawing a yellow SweetTart three times in a row if each SweetTart selected is eaten before the next is drawn". But there are only 2 yellow. So is there a mistake? Wait, no, let's check again. Wait, the counts: 4 pink, 1 blue, 2 yellow, 3 purple, 2 green. Total: 4+1=5, 5+2=7, 7+3=10, 10+2=12. So total 12. Yellow is 2. So to draw yellow three times, but we only have 2 yellow. So the probability should be 0? Wait, maybe the problem was supposed to be "drawing a yellow or maybe another color? Wait, no, the problem says "drawing a yellow SweetTart three times in a row". Since there are only 2 yellow, after drawing two, there are none left. So the probability of drawing three yellows in a row (without replacement) is 0, because we can't draw a third yellow when there are only two. Wait, but maybe I made a mistake in counting. Wait, 4 pink, 1 blue, 2 yellow, 3 purple, 2 green. 4+1=5, +2=7, +3=10, +2=12. Yes, total 12. Yellow: 2. So first draw yellow: 2/12. Second draw yellow: 1/11. Third draw yellow: 0/10. So multiply them: (2/12)(1/11)(0/10) = 0.

Wait, but maybe the problem was a typo, like maybe "drawing a pink" or another color, but as per the given problem, with 2 yellow, drawing three in a row (without replacement) is impossible, so probability 0.

Answer:

\( 0 \)