I need help on #3 I don't understand systems of linear inequalities since I wasn't there when my teacher explained it
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Answer:
The probability that she gets all the red ones, given that she gets the fluorescent pink one, is P=0.0035 or 0.35%.
Step-by-step explanation:
Susan grabs four marbles at random.
We have to calculate the probabilities that he picks the 3 red ones, given that she already picked the fluorescent pink.
If it is given that the fluorescent pink is already picked, we are left with three red marbles, four green ones, two yellow ones, and four orange ones. A total of 13 marbles.
The probability that the second marble is red is 3 in 13.
The probability that the third marble is also red is 2 (the red marbles that are left) in 12 (the total amount of marbles left), as there is a picking without replacement.
The probability that the fourth marble is 1 in 11.
Then, the probability that the 3 red marbles are picked, is:
x*y' + y = 8x
y' + y/x = 8 .... divide everything by x
dy/dx + y/x = 8
dy/dx + (1/x)*y = 8
We have something in the form
y' + P(x)*y = Q(x)
which is a first order ODE
The integrating factor is
Multiply both sides by the integrating factor (x) and we get the following:
dy/dx + (1/x)*y = 8
x*dy/dx + x*(1/x)*y = x*8
x*dy/dx + y = 8x
y + x*dy/dx = 8x
Note the left hand side is the result of using the product rule on xy. We technically didn't need the integrating factor since we already had the original equation in this format, but I wanted to use it anyway (since other ODE problems may not be as simple).
Since (xy)' turns into y + x*dy/dx, and vice versa, this means
y + x*dy/dx = 8x turns into (xy)' = 8x
Integrating both sides with respect to x leads to
xy = 4x^2 + C
y = (4x^2 + C)/x
y = (4x^2)/x + C/x
y = 4x + Cx^(-1)
where C is a constant. In this case, C = -5 leads to a solution
y = 4x - 5x^(-1)
you can check this answer by deriving both sides with respect to x
dy/dx = 4 + 5x^(-2)
Then plugging this along with y = 4x - 5x^(-1) into the ODE given, and you should find it satisfies that equation.