A bag contains 5 blue balls and 7 red balls. A ball is taken from the bag and its colour is noted. It is not replaced. A second ball is then taken from the bag and its colour is noted. Find the probability of obtaining: a) a blue ball and a red ball in either order

To solve this problem, we'll calculate the probability of obtaining one blue ball and one red ball in either order. The problem can be approached by considering the two possible orders:

1. **First a blue ball, then a red ball.**
2. **First a red ball, then a blue ball.**

### Step 1: Calculate the probability for each order.

**1. Probability of first drawing a blue ball and then a red ball:**

- Probability of drawing a blue ball first:  
  \[
  \text{P(Blue first)} = \frac{5}{12}
  \]
  (since there are 5 blue balls out of 12 total balls).

- After drawing the blue ball, there are now 11 balls left, with 7 of them being red.
  
  Probability of drawing a red ball next:
  \[
  \text{P(Red second | Blue first)} = \frac{7}{11}
  \]

- Therefore, the combined probability for this sequence is:
  \[
  \text{P(Blue, then Red)} = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}
  \]

**2. Probability of first drawing a red ball and then a blue ball:**

- Probability of drawing a red ball first:  
  \[
  \text{P(Red first)} = \frac{7}{12}
  \]
  (since there are 7 red balls out of 12 total balls).

- After drawing the red ball, there are now 11 balls left, with 5 of them being blue.

  Probability of drawing a blue ball next:
  \[
  \text{P(Blue second | Red first)} = \frac{5}{11}
  \]

- Therefore, the combined probability for this sequence is:
  \[
  \text{P(Red, then Blue)} = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}
  \]

### Step 2: Add the probabilities of the two possible sequences.

The total probability of obtaining one blue ball and one red ball in either order is:
\[
\text{Total Probability} = \text{P(Blue, then Red)} + \text{P(Red, then Blue)} = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}
\]

### Final Answer:

The probability of obtaining one blue ball and one red ball in either order is \(\frac{35}{66}\).

Would you like more details, or do you have any questions?

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Here are 5 related questions to explore further:

1. What is the probability of drawing two blue balls in a row?
2. What is the probability of drawing two red balls in a row?
3. How would the probability change if the balls were replaced after each draw?
4. What is the probability of drawing at least one red ball in two draws?
5. How would the probability change if there were 3 blue balls and 9 red balls instead?

**Tip:** When solving probability problems involving sequences of events, consider each event's individual probability and how they combine.

Explore the probability of drawing one blue ball and one red ball from a bag containing 5 blue balls and 7 red balls without replacement. Understand how to calculate probabilities using conditional probability and basic counting principles. Suitable for Grades 7-9 students.

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