vipulgoyal wrote:set S consists of numbers 2,3,6,48, and 164, Number K is computed by multiplying one rendom no from set S by one of the first 10 non negative integers, also selected at rendom , If Z=6^k , what is the probablity that 678463 is not a multiple of Z
Ans [spoiler]9/10[/spoiler]
P(678463 is NOT a multiple of Z) = 1 - P(678463 is a multiple of Z).
Options for Z:
An ODD integer cannot be a multiple of an EVEN integer.
Thus, 678463 will be a multiple of Z only if Z is odd.
Z = 6^k will be odd ONLY WHEN k=0, so that Z = k� = 1.
Thus, 678463 will be a multiple of Z only when k=0.
Options for k:
S = {2, 3, 6, 48, 164}
First 10 non-negative integers = {
0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
To determine the options for k, each value in S is to be multiplied by each of the 10 values in the second set.
Thus, for each value in S, there will be 10 products.
Of these 10 products, EXACTLY ONE will be equal to 0.
Thus, the probability that k=0 is 1/10.
Thus:
P(678463 is a multiple of Z) = 1/10.
Answer:
P(678463 is NOT a multiple of Z) = 1 - 1/10 = [spoiler]9/10[/spoiler].
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