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A class of twelve children has two more boys than girls. A group of three children are randomly picked from this class to accompany the teacher on a field trip. What is the probability that the group accompanying the teacher contains more girls than boys?
  • a)
    0
  • b)
    325/864
  • c)
    525/864
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A class of twelve children has two more boys than girls. A group of th...
There are 7B and 5G
Through as question is stating 3 students are taken at random.
This can be a possible way to approach it.

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A class of twelve children has two more boys than girls. A group of th...
To solve this problem, we need to first determine the total number of ways to choose a group of three children from the class of twelve. We can then calculate the number of ways to choose a group with more girls than boys and divide it by the total number of ways to get the probability.

Total Number of Ways to Choose a Group
To calculate the total number of ways to choose a group of three children from twelve, we can use the combination formula, which is given by nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items to be chosen.

In this case, we have twelve children and we want to choose a group of three. So, the total number of ways to choose the group is 12C3 = 12! / (3!(12-3)!) = 12! / (3!9!) = (12 * 11 * 10) / (3 * 2 * 1) = 220.

Number of Ways to Choose a Group with More Girls than Boys
Since there are two more boys than girls in the class, we can have 0, 1, or 2 girls in the group. Let's consider each case separately:

Case 1: 0 girls and 3 boys
To choose 3 boys from the 8 available, we can use the combination formula again. So, the number of ways to choose 3 boys is 8C3 = 8! / (3!(8-3)!) = (8 * 7 * 6) / (3 * 2 * 1) = 56.

Case 2: 1 girl and 2 boys
To choose 1 girl from the 4 available and 2 boys from the 8 available, we can again use the combination formula. So, the number of ways to choose 1 girl and 2 boys is 4C1 * 8C2 = (4! / (1!(4-1)!)) * (8! / (2!(8-2)!)) = (4 * 8 * 7) / (1 * 2 * 1) = 112.

Case 3: 2 girls and 1 boy
Similar to the previous cases, the number of ways to choose 2 girls and 1 boy is 4C2 * 8C1 = (4! / (2!(4-2)!)) * (8! / (1!(8-1)!)) = (4 * 3 / (2 * 1)) * 8 = 96.

Total Number of Ways to Choose a Group with More Girls than Boys
To get the total number of ways to choose a group with more girls than boys, we add the results from each case: 56 + 112 + 96 = 264.

Probability
Finally, we can calculate the probability by dividing the number of ways to choose a group with more girls than boys (264) by the total number of ways to choose a group (220): 264 / 220 = 33 / 55 = 3 / 5.

Therefore, the correct answer is option 'D' None of these.
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A class of twelve children has two more boys than girls. A group of three children are randomly picked from this class to accompany the teacher on a field trip. What is the probability that the group accompanying the teacher contains more girls than boys?a)0b)325/864c)525/864d)None of theseCorrect answer is option 'D'. Can you explain this answer?
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