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Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8
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19 Jul 2021, 02:31
19 Jul 2021, 02:31
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Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd?
(A) 1/9
(B) 1/6
(C) 2/9
(D) 1/4
(E) 1/2
(A) 1/9
(B) 1/6
(C) 2/9
(D) 1/4
(E) 1/2
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Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8
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19 Jul 2021, 11:02
19 Jul 2021, 11:02
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GeminiHeat wrote:
Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd?
(A) 1/9
(B) 1/6
(C) 2/9
(D) 1/4
(E) 1/2
(A) 1/9
(B) 1/6
(C) 2/9
(D) 1/4
(E) 1/2
Possible ways of picking 2 boys from 3 = \(^3C_2 = 3\)
Total ways of picking 2 girls from 3 = \(^3C_2 = 3\)
Total ways of picking 4 children = \((3)(3) = 9\)
The denominator must be a factor of 9 (1, 3, or 9)
Eliminate B, D and E straight away
Now, we have been asked that when 4 numbers are added - the result is ODD or EVEN!
Lets calculate the number of EVEN cases as B + B + G + G
2 Bs can add upto: 10 or 11 or 13
2Gs can add upto: 13 or 14 or 17
EVEN cases = (10 + 14), (11 + 13), (11 + 17), (13 + 13), and (13 + 17) i.e. 5 cases
So we must have 9 - 5 = 4 cases for ODD
Required difference = EVEN - ODD = \(\frac{5}{9} - \frac{4}{9} = \frac{1}{9}\)
Hence, option A
General Discussion
Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8
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31 Aug 2024, 20:38
31 Aug 2024, 20:38
1
Call Z_b the sum of the boys ages and Z_g the sum of the girls ages.
The set of the ages of the boys contains two even numbers and one odd number. In order for Z_b to be even, both of the selected ages must be even, so the probability of Z_b being even is the probability of not picking the odd number or 1/3. Thus, the probability of Z_b being odd is 2/3
The set of the ages of the girls contains two odd numbers and one even number. In order for Z_g to be even, both of the selected ages must be odd, so the probability of Z_g being even is the probability of not picking the even number, or 1/3. The probability of Z_g being odd is then 2/3.
Z can be even if both Z_g and Z_b are even or if Z_g and Z_b are odd. Thus, the probability of Z being even is:
P(Z - even) = P(Z_g - even) * P(Z_b - even) + P(Z_g - odd) * P(Z_b - odd)
P(Z - even) = 1/3 * 1/3 + 2/3 * 2/3 = 5/9
Then:
P(Z - odd) = 1 - P(Z - even) = 4/9
P(Z - even) - P(Z - odd) = 5/9 - 4/9 = 1/9.
The set of the ages of the boys contains two even numbers and one odd number. In order for Z_b to be even, both of the selected ages must be even, so the probability of Z_b being even is the probability of not picking the odd number or 1/3. Thus, the probability of Z_b being odd is 2/3
The set of the ages of the girls contains two odd numbers and one even number. In order for Z_g to be even, both of the selected ages must be odd, so the probability of Z_g being even is the probability of not picking the even number, or 1/3. The probability of Z_g being odd is then 2/3.
Z can be even if both Z_g and Z_b are even or if Z_g and Z_b are odd. Thus, the probability of Z being even is:
P(Z - even) = P(Z_g - even) * P(Z_b - even) + P(Z_g - odd) * P(Z_b - odd)
P(Z - even) = 1/3 * 1/3 + 2/3 * 2/3 = 5/9
Then:
P(Z - odd) = 1 - P(Z - even) = 4/9
P(Z - even) - P(Z - odd) = 5/9 - 4/9 = 1/9.
