Let's solve each part of the problem step by step.

(a) How many pairs are possible?

To find the total number of pairs that can be formed by selecting one student from each class, we simply multiply the number of students in the first class by the number of students in the second class.

• Number of students in the first class: 60
• Number of students in the second class: 50

$\text{Total number of pairs} = 60 \times 50 = 3000$

(b) What is the probability of one boy and one girl?

To calculate this probability, we consider the following cases:

• Selecting a boy from the first class and a girl from the second class.
• Selecting a girl from the first class and a boy from the second class.

Case 1: Selecting a boy from the first class and a girl from the second class:

• Number of boys in the first class: 30
• Number of girls in the second class: 20

The number of favorable outcomes for this case: $30 \times 20 = 600$

Case 2: Selecting a girl from the first class and a boy from the second class:

• Number of girls in the first class: $60 - 30 = 30$
• Number of boys in the second class: $50 - 20 = 30$

The number of favorable outcomes for this case: $30 \times 30 = 900$

So, the total number of favorable outcomes: $600 + 900 = 1500$

Now, the probability is: $\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{1500}{3000} = \frac{1}{2}$

(c) What is the probability of both being boys?

For both being boys:

• Number of boys in the first class: 30
• Number of boys in the second class: 30

The number of favorable outcomes: $30 \times 30 = 900$

Now, the probability is: $\text{Probability} = \frac{900}{3000} = \frac{3}{10}$

(d) What is the probability of at least one girl?

The probability of "at least one girl" is the complement of the probability of "no girls" (which means both selected students are boys).

From part (c), the probability of both being boys is $\frac{3}{10}$.

Thus, the probability of at least one girl is: $\text{Probability} = 1 - \frac{3}{10} = \frac{7}{10}$

Summary:

• (a) Number of pairs possible: 3000
• (b) Probability of one boy and one girl: $\frac{1}{2}$
• (c) Probability of both being boys: $\frac{3}{10}$
• (d) Probability of at least one girl: $\frac{7}{10}$

Would you like any details or have any questions?

Here are 5 related questions:

1. How would the probability change if the second class had 25 boys instead of 30?
2. What is the probability of selecting two girls, one from each class?
3. If one more boy is added to the first class, how does that affect the probabilities?
4. How would the results change if the first class had 70 students instead of 60?
5. What is the probability of selecting one boy and one girl if both classes had an equal number of students?

Tip: When solving probability problems, always consider whether you can simplify by using complementary probabilities, as it often makes calculations easier.