A class consists of 12 boys and 8 girls - Leaving Cert Mathematics - Question 1 - 2019

To calculate this probability, we can find the probability of selecting boys first and then a girl.

1. First Three are Boys:

• Probability of first boy: $\frac{12}{20}$
• Probability of second boy: $\frac{11}{19}$
• Probability of third boy: $\frac{10}{18}$

2. Fourth is a Girl:
   The probability of selecting a girl after selecting three boys: $\frac{8}{17}$

3. Total Probability:
   The combined probability is:
   $P(Boy, Boy, Boy, Girl) = \frac{12}{20} \times \frac{11}{19} \times \frac{10}{18} \times \frac{8}{17} = \frac{10560}{116280} = \frac{88}{969}.$

To find the total combinations of questions that can be answered, we consider the choices in both sections:

1. Section A Combinations:

• Question 1 must be answered. We then choose 3 questions out of the remaining 6 questions.
  Thus, the number of ways is:
  $C(6, 3) = \frac{6!}{3!(6-3)!} = 20.$

2. Section B Combinations:
   We must choose any 4 questions out of 8.
   So, the number of ways is:
   $C(8, 4) = \frac{8!}{4!(8-4)!} = 70.$

3. Total Combinations:
   The total number of combinations is the product of combinations from both sections:
   $Total = C(6, 3) \times C(8, 4) = 20 \times 70 = 1400.$