10/16 = .625
Analysis:
a and d will be odd if and only if both are odd. Therefor, ad will be odd 1/4 the time.
The same goes for bc. It will be odd 1/4 off the time.
ad-bc is even if and only if both ad and bc are either odd or even.
It's easiest to solve it from here with a table:
ad bc ad-bc prob
even even even 3/4*3/4 = 9/16
even odd odd 3/4*1/4 = 3/16
odd even odd 1/4*3/4 = 3/16
odd odd even 1/4*1/4 = 1/16
So, ad-bc will be even 9/16+1/16 = 10/16 of the time.
In the set of the first n integers, the number of a square number is approximately sqrt(n). So the probability of a square number is sqrt(n)/n = 1/sqrt(n). As n becomes larger this probability tends towards 0.
15/100=0.15
I will assume that you mean a five card poker hand. We can label the cards C1, C2, C3, C4, and C5. We are basically told already that C1 and C2 are both aces. So we have to find the probability that exactly one of C3, C4, and C5 is an ace. Knowing that the first two cards in our hand are both aces means that there are only 50 cards left in the deck. The probability that C3 is an ace and that C4 and C5 are both not aces is (2/50)(48/49)(47/48)=0.03836734694. The same probability also applies to each of C4 and C5, considered independently of each other. Therefore, our final probability is 3* 0.03836734694=0.1151020408
AnswerThe probability that a randomly chosen [counting] number is not divisible by 2 is (1-1/2) or 0.5. One out of two numbers is divisible by two, so 1-1/2 are not divisible by two.The probability that a randomly chosen [counting] number is not divisible by 3 is (1-1/3) = 2/3.Similarly, the probability that a randomly chosen [counting] number is not divisible by N is (1-1/N).The probability that a random number is not divisible by any of 2, 3 or 6 can be reduced to whether it is divisible by 2 or 3 (since any number divisible by 6 can definitely be divided by both and so it is irrelevant). This probability depends on the range of numbers available. For example, if the range is all whole numbers from 0 to 10 inclusive, the probability is 3/11, because only the integers 1, 5, and 7 in this range are not divisible by 2, 3, or 6. If the range is shortened, say just from 0 to 1, the probability is 1/2.Usually questions of this sort invite you to contemplate what happens as the sampling range gets bigger and bigger. For a very large range (consisting of all integers between two values), about half the numbers are divisible by two and half are not. Of those that are not, only about one third are divisible by 3; the other two-thirds are not. That leaves 2/3 * 1/2 = 1/3 of them all. As already remarked, a number not divisible by two and not divisible by three cannot be divisible by six, so we're done: the limiting probability equals 1/3. (This argument can be made rigorous by showing that the probability differs from 1/3 by an amount that is bounded by the reciprocal of the length of the range from which you are sampling. As the length grows arbitrarily large, its reciprocal goes to zero.)This is an example of the use of the inclusion-exclusion formula, which relates the probabilities of four events A, B, (AandB), and (AorB). It goes like this:P(AorB) = P(A) + P(B) - P(AandB)In this example, A is the event "divisible by 2", and B is the event "divisible by 3".
18 positive integers and 36 integers (negative and positive)
The sum of the integers from 1 to 100 inclusive is 5,050.
There are 1,000 positive integers between 1,000 and 9,999, inclusive, that are divisible by nine.
To be pedantic, it is not.
22.
2550
7
10
The sum of the positive integers from 1,000 to 1,100 inclusive is: 106,050
There are 22 integers between them.
The sum of all integers from 1 to 20 inclusive is 210.
It is 50.5
1,4,9,16,25,36,49,64,81