Hello,
Can you please assist with this problem:
If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
OA: [spoiler]10/21[/spoiler]
Positive divisors of 64 are 1, 2, 4, 8, 16, 32, 64
The following pairs have a sum which is less than 32:
1,2
1,4
1,8
1,16
2,4
2,8
2,16
4,8
4,16
So probability of selecting two distinct positive divisors whose sum is less than 32 is:
P(Selecting a 1 and 2 or 4 or 8 or 16) = 1/7 x 4/6 = 4/42
+
P(Selecting a 2 and 4 or 8 or 16) = 1/7 x 3/6 = 3/42
+
P(Selecting a 4 and 8 or 16) = 1/7 x 2/6 = 2/42
= 4/42 + 3/42 + 2/42
= 9/42
= 3/14
Can you please tell me where I am going wrong? Thanks for your help.
Best Regards,
Sri
Can you please assist with this problem:
If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
OA: [spoiler]10/21[/spoiler]
Positive divisors of 64 are 1, 2, 4, 8, 16, 32, 64
The following pairs have a sum which is less than 32:
1,2
1,4
1,8
1,16
2,4
2,8
2,16
4,8
4,16
So probability of selecting two distinct positive divisors whose sum is less than 32 is:
P(Selecting a 1 and 2 or 4 or 8 or 16) = 1/7 x 4/6 = 4/42
+
P(Selecting a 2 and 4 or 8 or 16) = 1/7 x 3/6 = 3/42
+
P(Selecting a 4 and 8 or 16) = 1/7 x 2/6 = 2/42
= 4/42 + 3/42 + 2/42
= 9/42
= 3/14
Can you please tell me where I am going wrong? Thanks for your help.
Best Regards,
Sri
