Let's solve the problem step by step:

Given:

• $P(\text{black}) = \frac{1}{4}$
• $P(\text{white}) = \frac{2}{3}$
• $P(\text{yellow}) = ?$

Part (a): Work out $P(\text{yellow})$

In a probability problem like this, the sum of the probabilities of all possible outcomes must equal 1. Therefore, we have:

$P(\text{black}) + P(\text{white}) + P(\text{yellow}) = 1$

Substituting the given probabilities:

$\frac{1}{4} + \frac{2}{3} + P(\text{yellow}) = 1$

First, find a common denominator for the fractions:

$\frac{1}{4} = \frac{3}{12}, \quad \frac{2}{3} = \frac{8}{12}$

Now, add the probabilities:

$\frac{3}{12} + \frac{8}{12} = \frac{11}{12}$

So,

$\frac{11}{12} + P(\text{yellow}) = 1$

Subtract $\frac{11}{12}$ from 1:

$P(\text{yellow}) = 1 - \frac{11}{12} = \frac{1}{12}$

Thus,

$P(\text{yellow}) = \frac{1}{12}$

Part (b): Suitable Number of Counters

Let the total number of counters be $N$. We know the probabilities for each color, so we can express the number of each color as a fraction of $N$:

• $\text{Number of black counters} = \frac{1}{4}N$
• $\text{Number of white counters} = \frac{2}{3}N$
• $\text{Number of yellow counters} = \frac{1}{12}N$

For these to be integers, $N$ must be a multiple of the least common multiple (LCM) of the denominators 4, 3, and 12.

• The LCM of 4, 3, and 12 is 12.

Thus, a suitable number for $N$ is 12.

Now, calculate the number of counters for each color:

• $\text{Black} = \frac{1}{4} \times 12 = 3$
• $\text{White} = \frac{2}{3} \times 12 = 8$
• $\text{Yellow} = \frac{1}{12} \times 12 = 1$

Final Answer:

• Part (a): $P(\text{yellow}) = \frac{1}{12}$
• Part (b): Suitable numbers of counters: 3 black, 8 white, and 1 yellow.

Reason: These numbers satisfy the given probabilities exactly when the total number of counters is 12.

Would you like any further details or explanations?

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Additional Questions:

1. How would the probabilities change if you doubled the total number of counters while keeping the same ratios?
2. Can you find another suitable total number of counters that would also work for these probabilities?
3. How would the problem change if $P(\text{yellow})$ was given and you had to find $P(\text{white})$?
4. How can the concept of expected value be applied to this problem?
5. What if one more color (e.g., red) was added to the bag, how would that affect the existing probabilities?

Tip: When dealing with probability problems, it’s often helpful to express everything in terms of a common denominator to simplify calculations.