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22 Apr 2023, 05:53
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Difficulty:
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Question Stats:
47% (03:00) correct
53%
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For a survey, a total of 11 persons are to be selected from 2 families having 4 members each and one family having 5 members, if at least 3 persons are to be selected from each family, what is the probability that all members of two families are selected?
A. \(\frac{5}{66}\)
B. \(\frac{5}{53}\)
C. \(\frac{5}{43}\)
D. \(\frac{5}{33}\)
E. \(\frac{5}{23}\)
A. \(\frac{5}{66}\)
B. \(\frac{5}{53}\)
C. \(\frac{5}{43}\)
D. \(\frac{5}{33}\)
E. \(\frac{5}{23}\)
22 Apr 2023, 07:15
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ChandlerBong
Given
- Family 1: 4 members
- Family 2: 4 members
- Family 3: 5 members
Question: We need to select 11 members, with a condition that at least 3 members should be from each family.
Let's first find the number of ways the selection can be done.
We can perform the selection in three ways -
Case 1: Select 4 members from Family 1 and Family 2, and select 3 members from Family 3
Case 2: Select 3 members from Family 1 and Family 2, and select 5 members from Family 3
Case 3: Between Family 1 and Family 2, select 4 members from one family and select 3 members from the other family, and select 4 members from Family 3
Case 1: Select 4 members from Family 1 and Family 2, and select 3 members from Family 3
The number of ways 4 members can be selected from Family 1 and Family 2 = \(^4C_4 * ^4C_4\) = 1
The number of ways 3 members can be selected from Family 3 = \(^5C_3\) = 10
Total = 10 * 1 = 10 ways
Case 2: Select 3 members from Family 1 and Family 2, and select 5 members from Family 3
The number of ways 3 members can be selected from Family 1 and Family 2 = \(^4C_3 * ^4C_3\) = 4 * 4 = 16
The number of ways 5 members can be selected from Family 3 = \(^5C_5\) = 1
Total = 16 * 1 = 16 ways
Case 3: Between Family 1 and Family 2, select 4 members from one family and select 3 members from the other family, and select 4 members from Family 3
Between Family 1 and Family 2, we can select the family from which three members will be selected in 2 ways (either we can choose Family 1 or we can choose Family 2).
From the chosen family (assume we pick family 1), we can select 3 out of four members in \(^4C_3\) ways = 4 ways
From the other family (family 2), we can select 4 out of four members in \(^4C_4\) ways = 1 way
From Family 3, we can select four out of five members in \(^5C_4\) ways = 5 ways.
Total = 2 * 4 * 1 * 5 = 40
Summing up the total from individual cases = Total number of cases = 40 + 16 + 10 = 66 ways.
Favorable cases: Only in case 1 we are selecting all the members from two families, i.e. family 1 and family 2.
Favorable number of cases = 10
Required Probability =\( \frac{10}{66} = \frac{5}{33}\)
Option D
16 May 2023, 11:49
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Aren't we supposed to pick 11 members from 13 members, so the total number of outcomes should be 13C11?
gmatophobia
