We are given a bag that contains balls of different colors:

• 10 white balls
• 12 blue balls
• 13 red balls
• 7 yellow balls
• 8 green wooden balls

A ball is selected without replacement (WOR), and the task is to find the probability of selecting one green ball and then one red ball, rounding the answer to four decimal places.

Step 1: Total number of balls

The total number of balls in the bag is: $10 + 12 + 13 + 7 + 8 = 50 \text{ balls}$

Step 2: Probability of selecting a green ball first

The probability of selecting one green ball (since there are 8 green balls) is: $P(\text{green first}) = \frac{8}{50}$

Step 3: Probability of selecting a red ball second

After removing one green ball, there are now $49$ balls left in the bag. The probability of selecting one red ball (since there are 13 red balls) is: $P(\text{red second}) = \frac{13}{49}$

Step 4: Combined probability

Since both events (selecting a green ball and then a red ball) are independent, we multiply the probabilities: $P(\text{green and red}) = \frac{8}{50} \times \frac{13}{49}$

Now, let's calculate: $P(\text{green and red}) = \frac{8 \times 13}{50 \times 49} = \frac{104}{2450} \approx 0.0424$

So, the probability of selecting one green ball and then one red ball is 0.0424, which corresponds to D).

Would you like more details or have any other questions?

Related Questions:

1. What is the probability of selecting two red balls in a row without replacement?
2. What is the probability of selecting a white ball followed by a blue ball?
3. How does the probability change if replacement is allowed?
4. What is the probability of selecting one yellow ball and one white ball?
5. If two balls are selected simultaneously, what is the probability that both are green?

Tip:

When calculating probabilities without replacement, always adjust the total number of objects in the denominator after each selection.