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Orlov [11]
3 years ago
6

There is a 85% probability that Jack's dad gives him permission to go to the movies tonight. There is a 60% chance that he will

be able to go to the movies and drive his car there alone. What is the probability that he will be able to drive to the movies alone given that his dad gives him permission to go?
Mathematics
1 answer:
kykrilka [37]3 years ago
5 0

Answer:

<u>The probability that Jack will be able to drive to the movies alone given that his dad gives him permission to go is 51%.</u>

Step-by-step explanation:

1. Let's check the information provided to us to answer the question correctly:

Probability that Jack's dad gives him permission to go to the movies tonight = 85% = 0.85

Probability that Jack will be able to go to the movies and drive his car there alone= 60% = 0.60

2. What is the probability that he will be able to drive to the movies alone given that his dad gives him permission to go?

For answering the question we will use the following formula for dependent events, given that Jack's father gave him permission to go:

Probability that Jack will be able to drive to the movies alone given that his dad gives him permission to go = Probability that Jack's dad gives him permission to go to the movies tonight * Probability that Jack will be able to go to the movies and drive his car there alone

Replacing with the real values, we have:

Probability that Jack will be able to drive to the movies alone given that his dad gives him permission to go = 85% * 60%

Probability that Jack will be able to drive to the movies alone given that his dad gives him permission to go = 0.85 * 0.60 = 0.51

<u>The probability that Jack will be able to drive to the movies alone given that his dad gives him permission to go is 51%.</u>

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Looks like f(x,y)=x^2+y^2-x^2y+7.

f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1

f_y=2y-x^2=0\implies2y=x^2

  • If x=0, then y=0 - critical point at (0, 0).
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The latter two critical points occur outside of D since |\pm\sqrt2|>1 so we ignore those points.

The Hessian matrix for this function is

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}

The value of its determinant at (0, 0) is \det H(0,0)=4>0, which means a minimum occurs at the point, and we have f(0,0)=7.

Now consider each boundary:

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f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4

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  • If x=-1, then

f(-1,y)=8-y+y^2

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

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f(x,1)=8

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f(x,-1)=2x^2+8

which has 3 extreme values, but the previous cases already include them.

Hence f(x,y) has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

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