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 find the total number of SweetTarts by adding 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 SweetTarts is 2. So the probability of drawing a yellow SweetTart first is \( P_1=\frac{2}{12}=\frac{1}{6} \).

Step3: Probability of second yellow (after eating first)

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

Step4: Probability of third yellow (after eating second)

After eating two yellow SweetTarts, the number of yellow SweetTarts 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, there is 1 yellow left and total 11. Third draw: after eating two yellows, there are 0 yellow left? Wait, no, the problem says "drawing a yellow SweetTart three times in a row". But we have only 2 yellow SweetTarts. Wait, maybe I made a mistake. Wait, let's re - check the numbers. Wait, the number of yellow is 2. So we can't draw three yellow in a row if we eat them, because there are only 2. Wait, maybe the problem has a typo? Wait, no, maybe I miscalculated the total. Wait, 4 (pink)+1(blue)+2(yellow)+3(purple)+2(green) = 4 + 1=5, 5+2 = 7, 7+3 = 10, 10+2 = 12. Correct. Number of yellow is 2. So to draw three yellow in a row, when we draw the first yellow (probability \( \frac{2}{12} \)), second yellow (probability \( \frac{1}{11} \)), and third yellow: but we only have 2 yellow, so after two draws, there are 0 yellow left. So the probability of drawing a third yellow is 0. So the combined probability is \( \frac{2}{12}\times\frac{1}{11}\times0 = 0 \). Wait, but maybe the problem meant without replacement? But with 2 yellow, we can't draw three. Alternatively, maybe the number of yellow is 3? Wait, the problem says 2 yellow. Let's check again. The problem states: 4 pink, 1 blue, 2 yellow, 3 purple, 2 green. So yellow is 2. So the probability of drawing three yellow in a row (without replacement, since we eat them) is 0, because we only have 2 yellow.

Wait, maybe I misread the problem. Let me check again. "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?" So since there are only 2 yellow SweetTarts, we can't draw three yellow ones in a row (because after two draws, there are no yellow left). So the probability is 0.

Alternatively, maybe the problem has a mistake, but based on the given numbers, let's proceed.

First, total number of SweetTarts: \( N=4 + 1+2 + 3+2=12 \)

Number of yellow: \( Y = 2 \)

Probability of first yellow: \( P_1=\frac{Y}{N}=\frac{2}{12}=\frac{1}{6} \)

After first yellow, number of yellow left: \( Y_1=2 - 1 = 1 \), total left: \( N_1=12 - 1 = 11 \)

Probability of second yellow: \( P_2=\frac{Y_1}{N_1}=\frac{1}{11} \)

After second yellow, number of yellow left: \( Y_2=1 - 1 = 0 \), total left: \( N_2=11 - 1 = 10 \)

Probability of third yellow: \( P_3=\frac{Y_2}{N_2}=\frac{0}{10}=0 \)

Then the probability of three yellow in a row is \( P = P_1\times P_2\times P_3=\frac{2}{12}\times\frac{1}{11}\times0 = 0 \)

Answer:

\( 0 \)