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A class consists of 12 boys and 8 girls - Leaving Cert Mathematics - Question 1 - 2019
Question 1
A class consists of 12 boys and 8 girls. What is the probability that the two students selected will be a boy and a girl in any order? Four students are selected, o... show full transcript
Worked Solution & Example Answer:A class consists of 12 boys and 8 girls - Leaving Cert Mathematics - Question 1 - 2019
Step 1
Two students are selected at random from the class. What is the probability that the two students selected will be a boy and a girl in any order?
Answer
To find the probability of selecting one boy and one girl in any order, we first calculate the total number of students:
Total students = 12 boys + 8 girls = 20 students.
We can select one boy and one girl in two ways:
- Boy first, then girl.
- Girl first, then boy.
Calculating the probabilities:
-
Probability (Boy first, then Girl):
P(Boy, Girl) = (\frac{12}{20} \times \frac{8}{19} = \frac{96}{380} = \frac{24}{95})
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Probability (Girl first, then Boy):
P(Girl, Boy) = (\frac{8}{20} \times \frac{12}{19} = \frac{96}{380} = \frac{24}{95})
Total Probability: [ P(Boy & Girl) = P(Boy, Girl) + P(Girl, Boy) = \frac{24}{95} + \frac{24}{95} = \frac{48}{95}]
Step 2
Four students are selected, one at a time, at random from the class. What is the probability that the first three students selected will be boys and the fourth will be a girl?
Answer
To find this probability, we first identify the probabilities for each selection:
- Probability of 1st boy = (\frac{12}{20})
- Probability of 2nd boy = (\frac{11}{19})
- Probability of 3rd boy = (\frac{10}{18})
- Probability of 4th girl = (\frac{8}{17})
Calculating the combined probability:
[ P(Boys, Boys, Boys, Girl) = \frac{12}{20} \times \frac{11}{19} \times \frac{10}{18} \times \frac{8}{17} ]
[ = \frac{10560}{116280} = \frac{88}{969}]
Step 3
Find how many different combinations of questions may be answered if a candidate follows this instruction.
Answer
For Section A:
- Must answer question 1 and select 3 from the remaining 6 questions.
- Combinations: (C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20)
For Section B:
- Need to select any 4 from 8 questions.
- Combinations: (C(8, 4) = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70)
Total combinations: [ Total = C(6, 3) \cdot C(8, 4) = 20 \cdot 70 = 1400]