The image contains a probability problem, and here's what it says:

1. Question: There is a bag containing 4 red balls and 3 green balls. Two balls are drawn consecutively without replacement. Determine the probability of:

   a. The first ball being red and the second being green.

   b. Both balls being green.

Solution:

a. Probability of the first ball being red and the second being green:

• Total number of balls: 4 red + 3 green = 7 balls.
• Probability of drawing a red ball first: $P(\text{Red}) = \frac{4}{7}$
• After drawing one red ball, 6 balls remain, and the probability of drawing a green ball second: $P(\text{Green | Red}) = \frac{3}{6} = \frac{1}{2}$
• Therefore, the combined probability is: $P(\text{Red first, Green second}) = \frac{4}{7} \times \frac{1}{2} = \frac{4}{14} = \frac{2}{7}$

b. Probability of both balls being green:

• The probability of drawing a green ball first: $P(\text{Green}) = \frac{3}{7}$
• After drawing one green ball, 6 balls remain, and the probability of drawing another green ball is: $P(\text{Green | Green}) = \frac{2}{6} = \frac{1}{3}$
• Therefore, the combined probability is: $P(\text{Both green}) = \frac{3}{7} \times \frac{1}{3} = \frac{3}{21} = \frac{1}{7}$

Summary of Answers:

• a. Probability of red first and green second: $\frac{2}{7}$
• b. Probability of both green: $\frac{1}{7}$

Do you want more details or have any questions?

Here are 5 related questions to explore further:

1. What would the probability be if the balls were drawn with replacement?
2. How does the probability change if the number of red and green balls is equal?
3. What is the probability of drawing two red balls consecutively without replacement?
4. How does the concept of conditional probability apply in this scenario?
5. Can you calculate the probability of drawing one ball of each color in any order?

Tip: When calculating probabilities without replacement, always remember to adjust the total number of objects after each draw.