To solve this problem, let's analyze the probability step-by-step.

Problem Breakdown:

• The ice chest contains:
  • 6 cans of apple juice
  • 8 cans of grape juice
  • 4 cans of orange juice
  • 2 cans of mango juice
• Total cans in the chest = $6 + 8 + 4 + 2 = 20$

Goal:

We want to find the probability of drawing three cans in a specific sequence:

1. A can of grape juice.
2. A can of apple juice.
3. A can of mango juice.

Solution:

This is a probability problem without replacement, so the total number of cans decreases with each draw.

1. Probability of first draw (Grape Juice):

   • There are 8 cans of grape juice out of 20 total cans.
   • Probability = $\frac{8}{20} = \frac{2}{5}$

2. Probability of second draw (Apple Juice):

   • After drawing one can of grape juice, 19 cans remain.
   • There are 6 cans of apple juice left.
   • Probability = $\frac{6}{19}$

3. Probability of third draw (Mango Juice):

   • After drawing a grape juice and an apple juice, 18 cans remain.
   • There are 2 cans of mango juice left.
   • Probability = $\frac{2}{18} = \frac{1}{9}$

Combined Probability:

The total probability of drawing a grape juice, then an apple juice, then a mango juice is the product of these probabilities:

$\frac{2}{5} \times \frac{6}{19} \times \frac{1}{9} = \frac{2 \times 6 \times 1}{5 \times 19 \times 9} = \frac{12}{855} = \frac{4}{285}$

Answer:

The probability is $\frac{4}{285}$.

Would you like more details on the calculation steps, or do you have further questions?

Related Questions:

1. What is the probability of drawing three cans in a row without any specific order?
2. How would the probability change if there were no mango juice in the ice chest?
3. What is the probability of drawing two cans of the same type in a row from this chest?
4. How does the probability calculation change if we replace each can after drawing it?
5. If we only wanted one specific juice, say grape, on the first draw, how would that probability look?

Tip:

When calculating probabilities with successive events, remember to adjust the total count for each draw if there’s no replacement.