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PS - Probability

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by tinni » Wed Jul 04, 2007 9:56 am
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA , after couple of discussions.
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Re: PS - Probability

by gabriel » Wed Jul 04, 2007 10:27 am
tinni wrote:If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA , after couple of discussions.


ok ... here goes .... in all n(n+1)(n+2) can have 96 different values ...

... for the condition given that n(n+1)(n+2) is divisible by 8 .. we can see that whenever n is a even number the function n(n+1)(n+2) is divisible by 8 ... also the function is divisible by 8 when (n+1) takes the value 8,16,24.... 96 ... so the total no. of values that the function is divisible by 8 is 48+12 = 60 .. so the probability is 60/96 = 5/8 ... so the answer is D ...
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Re: PS - Probability

by tinni » Wed Jul 04, 2007 10:48 am
Rock on !! Thank you.

I know where I went wrong. I haven't considered the (n+1) stuff, where we have 12 more numbers.
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