A bag contains 6 blue balls and 9 white balls. A ball is taken from the bag and its colour
noted. The ball 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 and a white ball in either order

To find the probability of obtaining one blue ball and one white ball in either order when drawing two balls without replacement, we can approach this problem by calculating the probability for both possible sequences and then summing these probabilities.

### Step 1: Total Balls in the Bag
Initially, there are a total of:
- 6 blue balls
- 9 white balls

So, there are \( 6 + 9 = 15 \) balls in the bag.

### Step 2: Possible Sequences
There are two possible sequences for drawing a blue and a white ball:
1. **Blue ball first, then white ball**
2. **White ball first, then blue ball**

### Case 1: Blue ball first, then white ball
1. **Probability of drawing a blue ball first:**
   \[
   P(\text{Blue first}) = \frac{6}{15}
   \]
2. **After drawing the blue ball, 14 balls remain (5 blue, 9 white). Probability of drawing a white ball next:**
   \[
   P(\text{White second}) = \frac{9}{14}
   \]
3. **Combined probability for this sequence:**
   \[
   P(\text{Blue first, White second}) = \frac{6}{15} \times \frac{9}{14} = \frac{54}{210} = \frac{9}{35}
   \]

### Case 2: White ball first, then blue ball
1. **Probability of drawing a white ball first:**
   \[
   P(\text{White first}) = \frac{9}{15} = \frac{3}{5}
   \]
2. **After drawing the white ball, 14 balls remain (6 blue, 8 white). Probability of drawing a blue ball next:**
   \[
   P(\text{Blue second}) = \frac{6}{14} = \frac{3}{7}
   \]
3. **Combined probability for this sequence:**
   \[
   P(\text{White first, Blue second}) = \frac{3}{5} \times \frac{3}{7} = \frac{9}{35}
   \]

### Step 3: Total Probability
Since the two sequences are mutually exclusive and cover all the possible ways to get one blue and one white ball, we add the probabilities:

\[
P(\text{Blue and White in either order}) = \frac{9}{35} + \frac{9}{35} = \frac{18}{35}
\]

### Final Answer
The probability of obtaining one blue ball and one white ball in either order is \(\boxed{\frac{18}{35}}\).

Would you like more details or have any other questions?

Here are 5 related questions you might find interesting:
1. What is the probability of drawing two blue balls in succession?
2. How would the probability change if the balls were replaced after each draw?
3. What is the probability of drawing two balls of the same color?
4. What would the probability be if there were an equal number of blue and white balls?
5. How does the probability change if a third ball is drawn without replacement?

**Tip:** When dealing with probabilities without replacement, always adjust the total number of possible outcomes after each event.

Calculate the probability of drawing one blue ball and one white ball in either order from a bag containing 6 blue balls and 9 white balls. Includes step-by-step solution with formulas and explanations. Suitable for Grades 9-12.

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