Chelsey puts twelve colored pieces of chocolate candy in a bag. Three pieces are green, four pieces are yellow, two pieces are purple, and three pieces are red. She draws one piece of candy at a time and eats it.

Question

What is the probability that she will draw out three pieces of candy in this order: red, green, yellow? (8 points; 2 per answer)

Red probability: 14 Green probability: 14 Yellow probability: 13

Multiply the probabilities to get an answer of: ??

What is the probability that she will draw out three pieces of candy in this order: green, yellow, red? (8 points; 2 per answer)

Green probability: 14 Yellow probability: 13 Red probability: 14

Multiply the probabilities to get an answer of: ??

What is the probability that she will draw out three pieces of candy in this order: yellow, red, green? (8 points; 2 per answer)

Yellow probability: ?? Red probability: ?? Green probability: ??

Multiply the probabilities to get an answer of: ??

Answer 1

To find the probabilities of drawing the candies in the specified orders, we will calculate each probability step-by-step.

1. Probability of drawing in the order: red, green, yellow

Total candies: 12 (3 green + 4 yellow + 2 purple + 3 red)

1. Red probability:

   • There are 3 red candies out of 12 total candies.
   • Probability = \( \frac{3}{12} = \frac{1}{4} \)

2. Green probability (after drawing red):

   • Now, there are 11 candies left, and still 3 green candies.
   • Probability = \( \frac{3}{11} \)

3. Yellow probability (after drawing red and green):

   • Now, there are 10 candies left, and still 4 yellow candies.
   • Probability = \( \frac{4}{10} = \frac{2}{5} \)

Total Probability: \[ \frac{1}{4} \times \frac{3}{11} \times \frac{2}{5} = \frac{6}{220} = \frac{3}{110} \]

2. Probability of drawing in the order: green, yellow, red

1. Green probability:

   • There are 3 green candies out of 12 total candies.
   • Probability = \( \frac{3}{12} = \frac{1}{4} \)

2. Yellow probability (after drawing green):

   • Now, there are 11 candies left, and still 4 yellow candies.
   • Probability = \( \frac{4}{11} \)

3. Red probability (after drawing green and yellow):

   • Now, there are 10 candies left, and still 3 red candies.
   • Probability = \( \frac{3}{10} \)

Total Probability: \[ \frac{1}{4} \times \frac{4}{11} \times \frac{3}{10} = \frac{12}{440} = \frac{3}{110} \]

3. Probability of drawing in the order: yellow, red, green

1. Yellow probability:

   • There are 4 yellow candies out of 12 total candies.
   • Probability = \( \frac{4}{12} = \frac{1}{3} \)

2. Red probability (after drawing yellow):

   • Now, there are 11 candies left, and still 3 red candies.
   • Probability = \( \frac{3}{11} \)

3. Green probability (after drawing yellow and red):

   • Now, there are 10 candies left, and still 3 green candies.
   • Probability = \( \frac{3}{10} \)

Total Probability: \[ \frac{1}{3} \times \frac{3}{11} \times \frac{3}{10} = \frac{9}{330} = \frac{3}{110} \]

Summary of answers

1. Probability for red, green, yellow: \( \frac{3}{110} \)
2. Probability for green, yellow, red: \( \frac{3}{110} \)
3. Probability for yellow, red, green: \( \frac{3}{110} \)

Final Answers

• For red, green, yellow:

  • Red probability: \( \frac{1}{4} \)
  • Green probability: \( \frac{3}{11} \)
  • Yellow probability: \( \frac{2}{5} \)
  • Multiply the probabilities: \( \frac{3}{110} \)

• For green, yellow, red:

  • Green probability: \( \frac{1}{4} \)
  • Yellow probability: \( \frac{4}{11} \)
  • Red probability: \( \frac{3}{10} \)
  • Multiply the probabilities: \( \frac{3}{110} \)

• For yellow, red, green:

  • Yellow probability: \( \frac{1}{3} \)
  • Red probability: \( \frac{3}{11} \)
  • Green probability: \( \frac{3}{10} \)
  • Multiply the probabilities: \( \frac{3}{110} \)