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18 Sep 2020, 23:41
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All divisors of 64: 1, 2, 4, 8, 16, 32, 64 = 7
The desired sum of two divisors less than 32: Sum of ant two factors among the first 5 will give the sum less than 32.
Total outcomes: \(^7{C_2}\) = 21
Desired outcome: \(^5{C_2}\) = 10
Probability: \(\frac{Desired outcome}{ Total outcome}\)
=> \(\frac{10 }{ 21}\)
Answer D
The desired sum of two divisors less than 32: Sum of ant two factors among the first 5 will give the sum less than 32.
Total outcomes: \(^7{C_2}\) = 21
Desired outcome: \(^5{C_2}\) = 10
Probability: \(\frac{Desired outcome}{ Total outcome}\)
=> \(\frac{10 }{ 21}\)
Answer D
19 Sep 2020, 08:41
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If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
divisors of 64 = 1,2,4,8,16, 32, 64
Pair of number whose sum is less than 32= (1,16) (1,8) (1,4) (1,2) (2,4) (2,8) (2,16) (4,8) (4,16) (8,16)= 10
Probability = No of pair whose sum is less than 32/Total number of outcome= 10/7C2=10/21
A. 1/3
B. 1/2
C. 5/7
D. 10/21
E. 15/28
divisors of 64 = 1,2,4,8,16, 32, 64
Pair of number whose sum is less than 32= (1,16) (1,8) (1,4) (1,2) (2,4) (2,8) (2,16) (4,8) (4,16) (8,16)= 10
Probability = No of pair whose sum is less than 32/Total number of outcome= 10/7C2=10/21
A. 1/3
B. 1/2
C. 5/7
D. 10/21
E. 15/28
12 Oct 2020, 06:14
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Divisors of 64: 1, 2, 4, 8, 16, 18, 32, 64
For the sum of any 2 factors to be more than 32, we need to have 32 or 64 as one of our factors.
P(not choosing 32 or 64) * P(not choosing 32 or 64 again)
5/7 * 4/6 = 20/42 = 10/21
For the sum of any 2 factors to be more than 32, we need to have 32 or 64 as one of our factors.
P(not choosing 32 or 64) * P(not choosing 32 or 64 again)
5/7 * 4/6 = 20/42 = 10/21
