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Number properties + probability

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Number properties + probability

by ellexay » Wed Feb 11, 2009 7:23 am
) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?



Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000
the first one is 8 and the last one is 1000
==> ((1000-8)/8) + 1 = 125
==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.
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by DanaJ » Wed Feb 11, 2009 7:57 am
I personally have no idea where they took that formula from. I know what i would do: divide 1000 by 8 and, after noticing that 1000 is a multiple of 8, then it is clear that the number of multiples of 8 is 1000/8 = 125. If you had smth that was not a multiple of 8 (say 1003), I'd divide this number by 8 as well and use the quotient for my answer.
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Re: Number properties + probability

by x2suresh » Wed Feb 11, 2009 9:25 am
ellexay wrote:) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?



Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000
the first one is 8 and the last one is 1000
==> ((1000-8)/8) + 1 = 125
==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.

first multiple of 8 between 1 and 1000 =8
last multiple of 8 between 1 and 1000 = 1000

total number of multiples = (1000-8)/8 = 124
Above forumula excluding number 8 itself..
So final answer = 124+1=125
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Re: Number properties + probability

by piyush_nitt » Thu Feb 12, 2009 4:36 am
ellexay wrote:) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?



Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000
the first one is 8 and the last one is 1000
==> ((1000-8)/8) + 1 = 125
==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.
AP formula

to determine the nth term

An = A+ (n-1)d

where An - nth Term
A - first term
n - number of terms
d - difference

In this case

1000 = 8 + (n-1)8

(1000-8) = (n-1)8

(1000-8)/8 + 1 = n
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