The image presents two probability problems involving the selection of balls from a box, both without and with replacement. Let's solve them one by one.

Problem 1 (Without Replacement):

There are 3 red balls and 5 green balls in a box. Two balls are drawn without replacement. We need to calculate two probabilities:

a) Probability that both balls are red:

• Probability of drawing the first red ball: $P(\text{first red}) = \frac{3}{8}$
• After drawing the first red ball, there are 2 red balls left and a total of 7 balls: $P(\text{second red}) = \frac{2}{7}$
• The combined probability (both events must happen) is the product: $P(\text{both red}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$

b) Probability that the first ball is red and the second is green:

• Probability of drawing the first red ball: $P(\text{first red}) = \frac{3}{8}$
• After drawing the first red ball, there are 5 green balls left and a total of 7 balls: $P(\text{second green}) = \frac{5}{7}$
• The combined probability is: $P(\text{first red, second green}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$

Problem 2 (With Replacement):

Here again, we have 3 red balls and 5 green balls in a box, but now two balls are drawn with replacement. This means after each draw, the ball is placed back into the box, so the total number of balls remains 8.

a) Probability that both balls are red:

• Probability of drawing the first red ball: $P(\text{first red}) = \frac{3}{8}$
• Since the ball is replaced, the probability of drawing a red ball again is the same: $P(\text{second red}) = \frac{3}{8}$
• The combined probability is: $P(\text{both red}) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{64}$

b) Probability that the first ball is red and the second is green:

• Probability of drawing the first red ball: $P(\text{first red}) = \frac{3}{8}$
• Since the ball is replaced, the probability of drawing a green ball is: $P(\text{second green}) = \frac{5}{8}$
• The combined probability is: $P(\text{first red, second green}) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}$

Would you like further details on any part of the solution or more examples?

Here are some related questions to think about:

1. What is the probability of drawing two green balls without replacement?
2. What is the probability of drawing two green balls with replacement?
3. How does replacement affect the total probabilities in general?
4. Can we calculate the probability of drawing one red and one green ball in any order?
5. What if there were more red or green balls—how would that change the outcomes?

Tip: When dealing with probability problems, always pay attention to whether the experiment is done with or without replacement—this changes the probabilities significantly!