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integer a,b,c,d not necessary distinct, are choosen independently and at random from the set
S={0,1,2,3,4,5........,2006,2007}. then find the probability that |ad-bc| is even
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There are an equal amount of odd and even numbers in the set S, and since a, b, c, and d do not need to be distinct we can simply treat all four as having a 50/50 chance of being either odd or even. Notice that |ad - bc| is even if ad and bc are either both odd or both even, and it is odd otherwise. Notice also that ad (and bc) is even if either a or d is even, and it is odd if both a and d are odd. So the probability that ad is odd is 1/2 * 1/2 = 1/4, and this goes for bc too. So the probability that ad and bc are both odd is 1/4 * 1/4 = 1/16, and the probability that both are even is 3/4 * 3/4 = 9/16, which gives us a 1/16 + 9/16 = 10/16 = 5/8 chance that |ad - bc| is even.
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thanks dude , now i have a doubt
whether 0 is even no. or odd no.
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0 is considered an even number.
Wrap it in bacon
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