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3 years ago

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Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

3 red, 2 blue, and 2 green balls are withdrawn; (b) at least 2 red balls are withdrawn; (c) all withdrawn balls are the same color; (d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.

1 answer:

UNO [17]3 years ago

8 0

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

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EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =

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$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

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Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

$\bigtriangledown f=\lambda \bigtriangledown g$

for some scalar $\lambda$ called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

$\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h$

The gradient of f is

$\bigtriangledown f=$

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in $x,y,z,\lambda,\mu$ .

We have

$f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0$

Let's compute the partial derivatives

$f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0$

The Lagrange condition leads to

$1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)$

Operating and simplifying

$1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11$

Replacing the value of $\lambda$ in the two first equations, we get

$1=11+2x\mu\\2=-11 +2y\mu$

From the first equation

$\displaystyle 2\mu=\frac{-10}{x}$

Replacing into the second

$\displaystyle 13=y\frac{-10}{x}$

Or, equivalently

$13x=-10y$

Squaring

$169x^2=100y^2$

To solve, we use the restriction h

$x^2 + y^2 = 1$

Multiplying by 100

$100x^2 + 100y^2 = 100$

Replacing the above condition

$100x^2 + 169x^2 = 100$

Solving for x

$\displaystyle x=\pm \frac{10}{\sqrt{269}}$

We compute the values of y by solving

$13x=-10y$

$\displaystyle y=-\frac{13x}{10}$

For

$\displaystyle x= \frac{10}{\sqrt{269}}$

$\displaystyle y= -\frac{13}{\sqrt{269}}$

And for

$\displaystyle x= -\frac{10}{\sqrt{269}}$

$\displaystyle y= \frac{13}{\sqrt{269}}$

Finally, we get z using the other restriction

$x - y + z = 1$

Or:

$z = 1-x+y$

The first solution yields to

$\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

And the second solution gives us

$\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Complete first solution:

$\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0

3 years ago