<span> 1.
If betting on red, 18/38 = 9/19 chance of winning, so 10/19 chance of losing
Chance of winning 4 times: (9/19)^4 = 6561/130321 =~ 5.034%
Chance of winning 3 times: (9/19)^3 * (10/19) * C(4,1) = 29160/130321 =~ 22.3755%
Chance of winning 2 times: (9/19)^2 * (10/19)^2 * C(4,2) = 48600/130321 =~ 37.2925%
Chance of winning 1 time: (9/19) * (10/19)^3 * C(4,3) = 36000/130321 =~ 27.624%
Chance of winning 0 times: (10/39)^4 = 10000/130321 =~ 76.7336% </span>
<span>The two points that are most distant from (-1,0) are
exactly (1/3, 4sqrt(2)/3) and (1/3, -4sqrt(2)/3)
approximately (0.3333333, 1.885618) and (0.3333333, -1.885618)
Rewriting to express Y as a function of X, we get
4x^2 + y^2 = 4
y^2 = 4 - 4x^2
y = +/- sqrt(4 - 4x^2)
So that indicates that the range of values for X is -1 to 1.
Also the range of values for Y is from -2 to 2.
Additionally, the ellipse is centered upon the origin and is symmetrical to both the X and Y axis.
So let's just look at the positive Y values and upon finding the maximum distance, simply reflect that point across the X axis. So
y = sqrt(4-4x^2)
distance is
sqrt((x + 1)^2 + sqrt(4-4x^2)^2)
=sqrt(x^2 + 2x + 1 + 4 - 4x^2)
=sqrt(-3x^2 + 2x + 5)
And to simplify things, the maximum distance will also have the maximum squared distance, so square the equation, giving
-3x^2 + 2x + 5
Now the maximum will happen where the first derivative is equal to 0, so calculate the first derivative.
d = -3x^2 + 2x + 5
d' = -6x + 2
And set d' to 0 and solve for x, so
0 = -6x + 2
-2 = -6x
1/3 = x
So the furthest point will be where X = 1/3. Calculate those points using (1) above.
y = +/- sqrt(4 - 4x^2)
y = +/- sqrt(4 - 4(1/3)^2)
y = +/- sqrt(4 - 4(1/9))
y = +/- sqrt(4 - 4/9)
y = +/- sqrt(3 5/9)
y = +/- sqrt(32)/sqrt(9)
y = +/- 4sqrt(2)/3
y is approximately +/- 1.885618</span>
Answer:
Step-by-step explanation:
Pretty sure is convergence lines
