2. A bag contains 6 blue and 9 yellow marbles.

For each condition below, find the probability of randomly selecting a blue marble on the first draw and a yellow marble on the second draw.



Answer :

Sure, let's go through the problem step by step to find the probabilities.

1. Calculate the total number of marbles initially:
- We have 6 blue marbles and 9 yellow marbles.
- Thus, the total number of marbles is:
[tex]\[ 6 + 9 = 15 \][/tex]

2. Calculate the probability of selecting a blue marble on the first draw:
- The number of blue marbles is 6.
- Therefore, the probability \( P(\text{Blue first}) \) is:
[tex]\[ P(\text{Blue first}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}} = \frac{6}{15} = 0.4 \][/tex]

3. Update the counts after drawing one blue marble:
- We now have 5 blue marbles left.
- The total number of marbles now is 14 (since one marble has been drawn).

4. Calculate the probability of selecting a yellow marble on the second draw:
- The number of yellow marbles remains 9 (since we only drew one blue marble first).
- Therefore, the probability \( P(\text{Yellow second | Blue first}) \) is:
[tex]\[ P(\text{Yellow second | Blue first}) = \frac{\text{Number of yellow marbles}}{\text{Remaining total number of marbles}} = \frac{9}{14} \approx 0.6428571428571429 \][/tex]

5. Calculate the combined probability of both events happening:
- We need the probability of drawing a blue marble first and then a yellow marble.
- The combined probability \( P(\text{Blue first and Yellow second}) \) is:
[tex]\[ P(\text{Blue first}) \times P(\text{Yellow second | Blue first}) = 0.4 \times 0.6428571428571429 = 0.2571428571428572 \][/tex]

So, the probabilities are as follows:
- Probability of drawing a blue marble on the first draw is \( 0.4 \).
- Probability of drawing a yellow marble on the second draw given a blue marble was drawn first is approximately \( 0.6428571428571429 \).
- The combined probability of this sequence of events is approximately \( 0.2571428571428572 \).

These steps should give you a comprehensive understanding of how to approach and solve such problems involving conditional probability.