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08 Feb 2021, 00:51
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D
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Difficulty:
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Question Stats:
65% (02:25) correct
35%
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wrong
based on 109
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Three groups of children contain respectively 3 girls and 1 boy; 2 girls and 2 boys; 1 girl and 3 boys. One child is selected at random from each group. What is the probability that the three selected children consists of one girl and 2 boys?
A. 7/32
B. 9/32
C. 11/32
D. 13/32
E. 15/32
Are You Up For the Challenge: 700 Level Questions
A. 7/32
B. 9/32
C. 11/32
D. 13/32
E. 15/32
Are You Up For the Challenge: 700 Level Questions
11 Feb 2021, 22:53
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Since the question mentions that there are three groups, and also mentions that a child is selected from each group at random, we have three possible combinations in which the children are selected.
Let's label the groups as Group A, Group B, Group C.
Combination 1: Girl (A), Boy (B), Boy (C)
Combination 2: Boy (A), Girl (B), Boy (C)
Combination 3: Boy (A), Boy (B), Girl (C)
Now, for each combination, we need to calculate the individual probability for each independent event (selection of girl or boy from the respective group)... and multiply them with each other to get a combined probability for that specific combination.
When we do that, we find:
Combination 1: P(Girl from A) * P(Boy from B) * P(Boy from C)
\(=\frac{ 3}{4} * \frac{2}{4} * \frac{3}{4} = \frac{18}{64}\)
Combination 2: P(Boy from A) * P(Girl from B) * P(Boy from C)
\(=\frac{1}{4} * \frac{2}{4} * \frac{3}{4} = \frac{6}{24}\)
Combination 3: P(Boy from A) * P(Boy from B) * P(Girl from C)
\(=\frac{1}{4} * \frac{2}{4} * \frac{1}{4} = \frac{2}{64}\)
When we add the combined possibilities for Combination 1, 2 and 3, we get:
\(\frac{18}{64}+\frac{6}{64}+\frac{2}{64} = \frac{26}{64} = \frac{13}{32}\) = Answer Choice D
Let's label the groups as Group A, Group B, Group C.
Combination 1: Girl (A), Boy (B), Boy (C)
Combination 2: Boy (A), Girl (B), Boy (C)
Combination 3: Boy (A), Boy (B), Girl (C)
Now, for each combination, we need to calculate the individual probability for each independent event (selection of girl or boy from the respective group)... and multiply them with each other to get a combined probability for that specific combination.
When we do that, we find:
Combination 1: P(Girl from A) * P(Boy from B) * P(Boy from C)
\(=\frac{ 3}{4} * \frac{2}{4} * \frac{3}{4} = \frac{18}{64}\)
Combination 2: P(Boy from A) * P(Girl from B) * P(Boy from C)
\(=\frac{1}{4} * \frac{2}{4} * \frac{3}{4} = \frac{6}{24}\)
Combination 3: P(Boy from A) * P(Boy from B) * P(Girl from C)
\(=\frac{1}{4} * \frac{2}{4} * \frac{1}{4} = \frac{2}{64}\)
When we add the combined possibilities for Combination 1, 2 and 3, we get:
\(\frac{18}{64}+\frac{6}{64}+\frac{2}{64} = \frac{26}{64} = \frac{13}{32}\) = Answer Choice D
27 Jul 2024, 13:17
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