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The coefficients of the quadratic equation x^2+bx+c=0 are determined by tossing a fair die twice, the first outcome to determine b and second to determine
c. what is the probability that the equation has real root (s), i.e., what is the probability that the discriminant b^2-4c ≥0 ?

Respuesta :

mathmate
This problem can be solved by a 36x36 contingency table.

The sample space of the sum of two throws of a die is 6*6=36.
We have the same sample space for the value of "b", and that of "c".

Let the rows of the 36x36 table represent b, and columns, c.

Calculate,  if b^2 ≥ 4c, mark a 1 in the square crossing the values of b & c.

At the end, count the number of "ones" in the table.

I get 986 ones out of 36^2=1296.
So the probability is 986/1296.

Alternatively, a simpler way is to calculate the number of occurrences of each outcome for two throws.

E.g. 
outcome freq 
2 1
3 2
4 3
5 4
6 5
7 6
8 5
9 4
10 3
11 2
12 1

Then we reduce the size of the table to 11x11, but we have to multiply the results by the weight (freq) to get the total probability.
facundo3141592

Here we found that the probability that the discriminant is larger than zero is equal to 0.5278

A fair die has 6 possible outcomes:

{1, 2, 3, 4, 5, 6}

So if we toss it two times, the total number of combinations will be equal to the product between the number of outcomes for each toss, this means that we have 6*6 = 36 possible combinations.

Now we need to find the combinations such that:

b^2 - 4c ≥0

So let's see the different values of b, and then let's see which values of c allow the above inequality to be true.

if b = 1, then we get:

1 - 4c ≥0  has no solutions, then we can discard b = 1.

if b = 2, then:

4 - 4c ≥0  has only one solution, c = 1

then the combination (b = 2, c = 1) makes the inequality true.

if b = 3 then:

9 - 4c ≥0  is true for c = 1 and c = 2

Then we have two combinations (b = 3,  c = 1) and (b = 3, c = 2)

if b = 4 then:

16 - 4c ≥0

here we have 4 solutions:

c = 1, c =2, c = 3 and c = 4.

Then we have 4 combinations here:

(b = 4, c  = 1), (b = 4, c  = 2), (b = 4, c  = 3), and (b = 4, c  = 4)

if b = 5, then:

25 - 4c ≥0

is true for any value of c, so here we have 6 combinations.

Trivially, the exact same thing happens for b  = 6, another 6 combination

Adding all the combinations such that b^2-4c ≥0 is true we got:

C = 0 + 1 + 2 + 4 + 6 + 6 = 19 combinations.

Then the probability that the discriminant will be equal or larger than zero is equal to the quotient between the number of combinations such that the inequality is true and the total number of combinations (36)

P = 19/36 = 0.5278

If you want to learn more about this subject, you can read:

https://brainly.com/question/24256398

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