Let's break down and solve the given probability problems step by step.

Given:

• 5 red balls
• 3 green balls
• Total balls in the bag = $5 + 3 = 8$

We are drawing two balls without replacement.

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(a) Probability of drawing a red ball followed by a green ball:

1. First draw (red ball):
   The probability of drawing a red ball on the first draw is: $P(\text{Red on 1st}) = \frac{5}{8}$

2. Second draw (green ball):
   After drawing one red ball, there are 7 balls left, 3 of which are green. So, the probability of drawing a green ball on the second draw is: $P(\text{Green on 2nd | Red on 1st}) = \frac{3}{7}$

3. Combined probability:
   Using the multiplication rule of independent events (because we are drawing without replacement), the probability of drawing a red ball followed by a green ball is: $P(\text{Red then Green}) = P(\text{Red on 1st}) \times P(\text{Green on 2nd | Red on 1st})$ $P(\text{Red then Green}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$

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(b) Probability of drawing two green balls:

1. First draw (green ball):
   The probability of drawing a green ball on the first draw is: $P(\text{Green on 1st}) = \frac{3}{8}$

2. Second draw (green ball):
   After drawing one green ball, there are 7 balls left, 2 of which are green. So, the probability of drawing another green ball is: $P(\text{Green on 2nd | Green on 1st}) = \frac{2}{7}$

3. Combined probability:
   The probability of drawing two green balls is: $P(\text{Green then Green}) = P(\text{Green on 1st}) \times P(\text{Green on 2nd | Green on 1st})$ $P(\text{Green then Green}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$

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(c) Probability that both balls are of the same color:

There are two possible outcomes for both balls being of the same color: both are red or both are green.

1. Probability of both red:
   The probability of drawing two red balls is calculated similarly: $P(\text{Red on 1st}) = \frac{5}{8}, \quad P(\text{Red on 2nd | Red on 1st}) = \frac{4}{7}$ $P(\text{Red then Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$

2. Probability of both green:
   We already calculated this in part (b): $P(\text{Green then Green}) = \frac{3}{28}$

3. Combined probability of both being the same color:
   Add the probabilities of both being red and both being green: $P(\text{Same color}) = P(\text{Red then Red}) + P(\text{Green then Green})$ $P(\text{Same color}) = \frac{20}{56} + \frac{3}{28} = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}$

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Final Answers:

(a) Probability of drawing a red ball followed by a green ball:
$P(\text{Red then Green}) = \frac{15}{56}$

(b) Probability of drawing two green balls:
$P(\text{Green then Green}) = \frac{3}{28}$

(c) Probability that both balls are of the same color:
$P(\text{Same color}) = \frac{13}{28}$

Would you like further details or clarifications?

Here are 5 related questions to think about:

1. What would be the probability of drawing a green ball followed by a red ball?
2. How would the probabilities change if you were drawing with replacement?
3. What is the probability of drawing one red and one green ball in any order?
4. What is the probability of drawing at least one green ball?
5. How would the results change if there were an equal number of red and green balls?

Tip: When dealing with probabilities involving "without replacement," always account for the fact that the total number of items decreases after each selection.