To solve this, we'll calculate the probability that Miss Pat randomly selects one boy and one girl when choosing two children from a group of 9 boys and 8 girls.

Total number of children:

There are $9$ boys and $8$ girls, making a total of: $9 + 8 = 17 \text{ children.}$

Total number of ways to select 2 children from 17:

The number of ways to choose 2 children from 17 is given by the combination formula: $\binom{17}{2} = \frac{17 \times 16}{2} = 136$

Number of ways to select one boy and one girl:

• Number of ways to select 1 boy from 9 boys: $\binom{9}{1} = 9$
• Number of ways to select 1 girl from 8 girls: $\binom{8}{1} = 8$

So, the total number of ways to select one boy and one girl is: $9 \times 8 = 72$

Probability:

The probability of selecting one boy and one girl is the ratio of favorable outcomes (one boy and one girl) to the total possible outcomes (any two children): $\text{Probability} = \frac{72}{136} = \frac{9}{17}$

Thus, the probability that one child is a girl and the other is a boy is: $\frac{9}{17}$

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

Here are five related questions to extend your understanding:

1. What is the probability of selecting two boys?
2. What is the probability of selecting two girls?
3. How would the probability change if there were 10 boys instead of 9?
4. What is the total number of ways to select 3 children from the 17?
5. If Miss Pat selects 4 children, what is the probability exactly 2 will be boys?

Tip: When calculating probabilities involving combinations, make sure to identify favorable outcomes and total possible outcomes clearly.