probability and statistic

Question

In an inequality a>=b the variables a and b are replaced by randomly selected single-digit natural numbers. Find the probability that the resulting inequality is correct.

Answer 1

Visualize a 10x10 table with numbers 0 through 9 along the top-header, representing numbers for "a."

Along the side-header, there are also numbers 0 through 9, representing numbers for "b."

There are 100 spaces inside the table. The top-left square represents (a,b) = (0,0) and the square to the right represents (a,b) = (1,0), and so on. Each of the 100 squares represents an ordered pair of (a,b).

Along the diagonal, the (a,b) pairs contain pairs where a=b.

The North-East triangle contains (a,b) pairs where a > b.

The number of pairs that are good (what we are looking for) where a >= b is 55.

To get that 55-number, count all the squares in the North-East triangle (above the diagonal) and add the number of squares in the diagonal. You will get 55 squares where a >=b.

Since there are 100 squares in total, and 55 squares represent what we are looking for, the probability of randomly selecting a square with a >=b is 55/100.

Or we could say: P(a>=b) = 55/100 = .55

EXTRA: Clever way to count the squares where a >= b.

Realize that if you cut the 10x10 square in half, via a line down the diagonal, you will cut each diagonal square in half. Half of 100 is 50. So that cut, down the diagonal gives us 50. Included in that 50 are half of the 10 diagonal squares, which represents 5 of our 50. (because 10 halves is 5).

But, we want all 10 diagonal squares to be included with the good squares, not just half of them.

The diagonal line, cut through the diagonal squares leaves 10-halves above, and 10-halves below.

If we add those 10-halves that were cut from the diagonal, we add 5 to get a total of 55.

If this is too abstract, just sketch the table and count the good squares!

Answer 2

To answer this question, we can break it down into multiple cases. We will first calculate the conditional probability that a>=b for specific values of b. Then we will use the multiplication rule and the law of total probability to determine the final answer as follows:

1. P(a>=b | b=0) = 1.0, P(b=0) = 0.1 -> P(a>=b AND b=0) = (1.0)(0.1) = 0.10
2. P(a>=b | b=1) = 0.9, P(b=1) = 0.1 -> P(a>=b AND b=1) = (0.9)(0.1) = 0.09
3. P(a>=b | b=2) = 0.8, P(b=2) = 0.1 -> P(a>=b AND b=2) = (0.8)(0.1) = 0.08
4. P(a>=b | b=3) = 0.7, P(b=3) = 0.1 -> P(a>=b AND b=3) = (0.7)(0.1) = 0.07
5. P(a>=b | b=4) = 0.6, P(b=4) = 0.1 -> P(a>=b AND b=4) = (0.6)(0.1) = 0.06
6. P(a>=b | b=5) = 0.5, P(b=5) = 0.1 -> P(a>=b AND b=5) = (0.5)(0.1) = 0.05
7. P(a>=b | b=6) = 0.4, P(b=6) = 0.1 -> P(a>=b AND b=6) = (0.4)(0.1) = 0.04
8. P(a>=b | b=7) = 0.3, P(b=7) = 0.1 -> P(a>=b AND b=7) = (0.3)(0.1) = 0.03
9. P(a>=b | b=8) = 0.2, P(b=8) = 0.1 -> P(a>=b AND b=8) = (0.2)(0.1) = 0.02
10. P(a>=b | b=9) = 0.1, P(b=9) = 0.1 -> P(a>=b AND b=9) = (0.1)(0.1) = 0.01

So P(a>=b) = P(a>=b AND b=0) + P(a>=b AND b=1) + ... + P(a>=b AND b=9) = 0.10 + 0.09 + ... + 0.01 = 0.55