really hard math questions that are insane!!! help!?

Question

ok! these are realy really hard, so please help me!

1) intergers 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?

2) How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

p.s. i have the answers, but not the how or the why, so please make sure that your answers are actually correct, because last time most of the answers i received were wrong, thanks ^^
answer 1: 5/8 answer 2: 112
but i really don't know how to do these! the answers were given to me because it was a multiply choice thing and i was told the correct ones after the test. ^^ HELP!

Answer 1

1)probability that ad-bc is even
if ad is odd and bc is odd
if ad is even and bc is even

a)ad is odd iff a is odd and d is odd
bc is odd iff a is odd and c is odd

b) ad is even => a is even , d is odd or d is even a is odd or both are even

c) bc is even => b is even , c is odd or c is even b is odd or both are even

There are 16 combinations possible (a,b,c,d) in terms of even and odd .

let us call odd as o and even as e : Then our set is

(o,o,o,o) ,(e,e,e,e),(o,e,o,e),(e,o,e,o),(o,o,e,e),(e,e,o,o)
,(o,e,e,e),(e,o,e,e),(e,e,o,e),(e,e,e,e)
Total no of combinations = 10

Probability = 10/16 = 5/8

2)

Answer 2

1) you have the same number of even and odd numbers in the range 0 to 2007. ( 0, 2, 4... and 1, 3, 5. It starts with an even number and ends with an odd number.) The proababilty each of the integers is even is .5.

even * even = even
even * odd = even
odd * even = even
odd * odd = odd

Therefore the probability that ab is even is .75 and the probability of bc is even is .75 or 3/4. Probability of an odd is 1/4.

odd - odd = even ==> 1/4 * 1/4 = 1/16
odd - even = odd ==> 1/4 * 3/4 = 3/16
even - odd = odd ==> 3/4 * 1/4 = 3/16
even - even = even ==> 3/4 * 3/4 = 9/16

So the probability of an even outcome is 1/16 + 9/16 = 10/16 or 5/8 = .625. Probabilities are often expressed as percentages between 0 and 1 inclusive.

2) The first digit can be anything, the second digit would be Ok 9 out of 10 times, the third digit will be OK 8 out of 10 times. So we have 1 * 9/10 * 8/10 = .72 probability that the digits are distinct (non-repeating). But then it gets tricky, so let's try a different approach. Here is a totally different approach:

List all of the combinations of numbers where we have the average represented by one of the digits.

0 1 2
0 2 4
0 3 6
0 4 8
1 2 3
1 3 5
1 4 7
1 5 9
2 3 4
2 4 6
2 5 8
3 4 5
3 5 7
3 6 9
4 5 6
4 6 8
5 6 7
5 7 9
6 7 8
7 8 9

So we have 20 combinations of numbers, but they could appear in a different order. For example 012 could be any of the following six combinations: (0 1 2), (0 2 1), ... and so on. In other words there are 6 X 20 or 120 different three digit combinations that match this criteria.

As an earlier poster said if you eliminate the 0 then you come up with the answer you want.

Answer 3

1. You can simplify number by using just the numbers 0 and 1 instead of a range of digits. 0 and 1 behave as any odd or even number. Of course there are 8 ways to pick 4 numbers out of 0 and 1. Using 0s and 1s each product can only be 0 or 1 with only one way to get a 1 (1*1). So there is a 3/4*3/4 chance of 0-0 = 0 = even, a 3/4*1/4 chance of 0 - 1 = -1 odd, a 1/4*3/4 chance of 1 - 0 = 1 odd and a 1/4*1/4 chance of 1 - 1 = 0 even. So the chance of an even result is 3/4*3/4 + 1/4*1/4 = 5/8


2. You have 3 distinct digits, meaning no digit is repeated. The average digit means that one digit is in between the other two. If the average digit is some number n than the other two digits must have the form n + m and n - m to average with m > 0.

In general, m <= n because if you subtract more than n, you go below zero. Similarly m + n <= 10 so m <= 10 - n

Now look at each possible average digit and see what the possible values of m are.

Of course, 0 there is no suitable m.
For 1, m = 1
For 2, m = 1, 2
For 3, m = 1, 2, 3
For 4, m = 1, 2, 3, 4
For 5, m = 1, 2, 3, 4
For 6, m = 1, 2, 3
For 7, m = 1, 2,
For 8, m = 1,
For 9, m = 1

So all the posibilities are just 2 x ( 1 + 2 + 3 + 4) = 25

So there are 20 3-digit combinations that fit the average condition. Of course each 3 digit combination can be arranged 6 ways so there are 6 x 20 = 120 possible combinatons. This counts combinations with leading zeros such as 012 which isn't really a 3 digit number. So how many leading zeros are there? Zeros happen whenever n = m because n - m then gives a zero digit. There are 4 of these cases. Each one gives 2 numbers with a leading zero for 8 total leaving 112 numbers that fit your condition.

Answer 4

Interesting problems. Hope this helps you understand the answers.

1) If you choose an integer at random from 0 to 2007, there is a 50% chance that it will be even. If a or d is even then their product is even. In order to get an odd product, both a and d have to be odd.That happens only 1 time in 4 (50% times 50%). The other 3 times out of 4, the product is even.

And the same is true of the other product, bc. (25% of the time it is odd; 75% of the time it's even.)

Now if both products are even, their difference is even. And if both products are odd, their difference is even.

The probability that ad and bc are both even is 3/4 times 3/4, which is 9/16. And the probability that both are odd is 1/4 times 1/4, which is 1/16. So the total probability that the difference is even is 9/16 + 1/16 = 10/16 = 5/8.

2) Count the possibilities as follows:

If the lowest of the 3 digits is 0, the largest one has to be 2, 4, 6, or 8. This is because the difference between largest and smallest has to be even; otherwise, the average is not an integer. If the lowest digit is 1, the largest has to be 3, 5, 7, or 9. So the number of combinations with either 0 or 1 as the lowest digit is 4 (for a total of 4+4 = 8).

By similar reasoning, there are 3 combinations each that have 2 or 3 as the lowest digit, for a total of 6 possibilities.

With 4 or 5 as the lowest digit, we get a total of 4 more possibilities. And with 6 or 7 as the lowest digit, there are just 2 more possibilities.

So in total there are 2+4+6+8 possibilities, which totals 20.

And for each combination of 3 numbers, there are 6 arrangements (e.g., for 1, 2, and 3, we have 123, 132, 213, 231, 312, 321). That would give us 6 x 20 = 120 different 3-digit numbers. But the ones that start with 0 don't count, because they are really 2-digit numbers.

How many of the number start with 0? Well, as mentioned earlier, there are 4 possible values for the largest of the digits that go with 0 (2, 4, 6, or 8). And for each of those, there is 1 other number that works (because it is the average of 0 and the other number). Of the six different arrangements for each set of numbers, 2 of these arrangements put the 0 first, so they don't satisfy the requirements. 4 digits (2,4,6,8) times 2 unacceptable arrangements each equals 8 numbers that have to be eliminated.

120 - 8 = 112, which is the final answer.

Answer 5

For 1), there are two ways that ad-bc can be even. Either both ad and bc are even, or both ad and bc are odd.

In order for ad to be even, then either a and d are both even (1/4 chance), or one of them is even and the other is odd (1/2 chance). So, there is a 3/4 chance of ad being even. Within this 3/4 chance, using the same logic, there's also a 3/4 chance that bc is even. So 3/4 x 3/4 = 9/16 chance that both ad and bc are even.

In order for ad to be odd, then both a and d must be odd. There is 1/4 chance that a and d are both odd. Within this 1/4 chance, there is a 1/4 chance that b and c are both odd. So 1/4 x 1/4 = 1/16 chance that both ad and bc are odd.

Now, add the two cases together: ad and bc both even is 9/16 chance, ad and bc both odd is 1/16 chance, 9/16+1/16=10/16=5/8

Oh yeah, 0-2007 doesn't really matter, as long as there is an equal number of odd and even numbers (0 counts as even). It may as well be a number 0-1 or 1-2 the result is the same.

Answer 6

1)
Between 0 and 2007, there are exactly 1003 odd integers, and 1003 even integers. Thus, each variable is going to be even half the time and odd half the time.
This means, the product of 2 of the variables is going to be even 3/4s of the time (even * even, even * odd, odd * even) and odd 1/4 of the time (odd * odd).
For ad-bc to be even, we need either an odd product minus an odd product or an even product minus an even product.
So the answer is 3/4 * 3/4 + 1/4 * 1/4
=9/16 + 1/16
=10/16
=5/8

2)
Possibilities for the digits:
012... 024... 036... 048... 123... 135... 147... 159... 234... 246... 258... 345... 357... 369... 456... 468... 567... 579... 678... 789
There are 20 different possible sets of 3 digits. Pretend they can all be arranged six ways (abc, acb, bac, bca, cab, cba). Then, we have 120 possible three-digit numbers. However, numbers that start with zero would be considered two-digit numbers, instead of three. Thus, for 012, 024, 036, and 048, 2 of the solutions are not valid (abc and acb).
This gives us 120 - 2 * 4 = 112

Answer 7

You mean a*d - b*c ?
First you determine what is the probability that a*d is even.
There are 4 cases
1. a and d are both even, then a*d is even
2. a and d are both odd, then a*d is odd
3. a is even, d is odd even, then a*d is even
4 a is odd, d is even even, then a*d is even

So there is a probability of 3/4 that a*d is even.
Same is true for b*c of course.

a*d - b*c is even if either a*d is even and b*c is even too
or if both products are odd.
even - even = even
odd - odd = even
even - odd = odd
odd - even = odd

So the probability for the result to be even is 3/4 * 3/4 + 1/4 * 1/4 = 10/16 = 5/8

Answer 8

I can only give you hints, but will not calculate for you.

1. Probability of each of a, b, c, d being odd is 0.5, same of them being even.
2. Even x even = even
Odd x even = even
odd x odd =odd therefore probability of each of ad and bc being odd is 0.5 x 0.5 = 0.25
that means probability of each of ad and bc being even is 1 - 0.5 = 0.75
3. odd - odd = even
even - even = even
odd - even = odd
4. possibilities
ad is even and bc is odd - the answer is odd
ad is even and bc is even - the answer is even
ad is odd and bc is odd - the answer is even
ad is odd and bc is even.- the answer is odd