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Integers a, b, c and d, not necessarily distinct, are chosen

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sandra123
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Integers a, b, c and d, not necessarily distinct, are chosen [#permalink] New post   25 Sep 2013, 04:20
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Integers a, b, c and d, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that ad - bc is even?

A. 3/8
B. 7/16
C. 1/2
D. 9/16
E. 5/8
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Bunuel
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Re: Integers a, b, c and d, not necessarily distinct, are chosen [#permalink] New post   25 Sep 2013, 04:37
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sandra123 wrote:
Integers a, b, c and d, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that ad - bc is even?

A. 3/8
B. 7/16
C. 1/2
D. 9/16
E. 5/8


First of all notice that half of the numbers from from 0 to 2007, inclusive are odd and half even.

Next, let's find the probability of ad - bc being odd and subtract it from 1.

The difference of two integers to be odd we should have either ODD-EVEN OR EVEN-ODD:

ODD-EVEN = ad - bc = odd*odd-even*even or odd*odd-odd*even or odd*odd-even*odd. Three cases. The probability of each is 1/2^4, so the probability of 3 cases is 3/2^4.

The same for EVEN-ODD: p=3/2^4.

P(even) = 1 - P(odd) = 1 - (3/2^4 + 3/2^4) = 10/16 = 5/8.

Answer: E.

Hope it's clear.
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fnumiamisburg
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Re: Integers a, b, c and d, not necessarily distinct, are chosen [#permalink] New post   25 Sep 2013, 08:19
Hello Bunel, I tried to approach the question in similar way but got it wrong.. :(
Can u plz explain hw we got the probability of each is 1/2^4
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Re: Integers a, b, c and d, not necessarily distinct, are chosen [#permalink] New post   25 Sep 2013, 08:37
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fnumiamisburg wrote:
Hello Bunel, I tried to approach the question in similar way but got it wrong.. :(
Can u plz explain hw we got the probability of each is 1/2^4


Consider one case, say ad - bc = odd*odd - even*even. What is the probability that a=odd, d=odd, b=even, and c=even? The probability that a=odd is 1/2, the probability that d=odd is 1/2, the probability that b=even is 1/2, and the probability that c=even is 1/2. Thus the probability that (a=odd, d=odd, b=even, and c=even)=1/2*1/2*1/2*1/2=1/2^4.

The same applies to other cases.

Hope it's clear.
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fnumiamisburg
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Re: Integers a, b, c and d, not necessarily distinct, are chosen [#permalink] New post   25 Sep 2013, 10:27
Oh!!!! yes..its clear now...it ws really simple..
Thanks
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Re: Integers a, b, c and d, not necessarily distinct, are chosen [#permalink] New post   12 Sep 2019, 05:44
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Re: Integers a, b, c and d, not necessarily distinct, are chosen [#permalink]
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