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07 Apr 2022, 03:15
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E
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Difficulty:
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86% (01:33) correct
14%
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wrong
based on 83
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In a group A, there are 7 boys and 3 girls, whilst in another group B, there are 2 boys and 8 girls. One person is selected at random from each group. What is the probability that one is a boy and one is a girl?
A. 7/50
B. 11/45
C. 19/50
D. 37/90
E. 31/50
A. 7/50
B. 11/45
C. 19/50
D. 37/90
E. 31/50
11 Apr 2022, 12:03
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Probability of an event E, P (E) =\( \frac{{Number of outcomes favourable to the event} }{ {Total possible outcomes of the random experiment}}\).
In this question, the random experiment being conducted is the selection of one person from each group.
Selecting any 1 person from group A can be done in \(10_C_1\) ways i.e. 10 ways (since \(n_C_1\) = n)
Selecting any 1 person from group B can be done in \(10_C_1\) ways i.e. 10 ways (since \(n_C_1\) = n)
Therefore, total possible outcomes = 10 * 10 = 100
The event for which the probability is to be found is the event that one of the selected persons is a boy and the other is a girl.
Visualising the scenario, the boy could be from group A and the girl from group B OR the other way round.
If the boy is from group A and the girl from group B, total number of such selections = \(7_C_1\) * \(8_C_1\) = 7 * 8 = 56
If the boy is from group B and the girl from group A, total number of such selections = \(2_C_1\) * \(3_C_1\) = 2 * 3 = 6
Number of outcomes favourable to the event = 56 + 6 = 62.
Therefore, required probability = \(\frac{62 }{ 100}\) = \(\frac{31 }{ 50}\).
The correct answer option is E.
In this question, the random experiment being conducted is the selection of one person from each group.
Selecting any 1 person from group A can be done in \(10_C_1\) ways i.e. 10 ways (since \(n_C_1\) = n)
Selecting any 1 person from group B can be done in \(10_C_1\) ways i.e. 10 ways (since \(n_C_1\) = n)
Therefore, total possible outcomes = 10 * 10 = 100
The event for which the probability is to be found is the event that one of the selected persons is a boy and the other is a girl.
Visualising the scenario, the boy could be from group A and the girl from group B OR the other way round.
If the boy is from group A and the girl from group B, total number of such selections = \(7_C_1\) * \(8_C_1\) = 7 * 8 = 56
If the boy is from group B and the girl from group A, total number of such selections = \(2_C_1\) * \(3_C_1\) = 2 * 3 = 6
Number of outcomes favourable to the event = 56 + 6 = 62.
Therefore, required probability = \(\frac{62 }{ 100}\) = \(\frac{31 }{ 50}\).
The correct answer option is E.
21 Oct 2024, 08:06
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