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2 years ago

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2.82 For married couples living in a certain suburb, the probability that the husband will vote on a bond referendum is 0.21, th

e probability that the wife will vote on the referendum is 0.28, and the probability that both the husband and the wife will vote is 0.15. What is the probability that (a) at least one member of a married couple will vote? (b) a wife will vote, given that her husband will vote? (c) a husband will vote, given that his wife will not vote?

1 answer:

skad [1K]2 years ago

4 0

Answer:

a) 0.3400

b) 0.7143

c) 0.0833

Step-by-step explanation:

Let H be the event that the husband will vote on the bond referendum, and W be the event that the wife will vote on the bond referendum.

P(H) = 0.21

P(W) = 0.28

P(H and W) = 0.15

a) The probability that at least one member of the couple will vote is:

$P(A\cup B) = P(A)+ P(B) - P(A\cap B)\\P(A\cup B) =0.21+0.28-0.15 = 0.34$

b) The probability that a wife will vote, given that her husband will vote is:

$P(W|H)=\frac{P(H\cap W)}{P(H)}=\frac{0.15}{0.21} \\P(W|H)=0.7143$

c) The probability that a husband will vote, given that his wife will not vote (W') is:

$P(H|W')=\frac{P(H) -P(H\cap W)}{1-P(W)}=\frac{0.21-0.15}{1-0.28} \\P(H|W')=0.0833$

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Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

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The solution is "The maximum number of solutions is one."

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The equation for the parabola without further analysis is $y=ax^2+9$ .

We are going to plug $ax^2+9$ into $x^2+y^2=r^2$ for $y$ .

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$x^2+(ax^2+9)^2=r^2$

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$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

$x^2+a^2x^4+18ax^2+81=r^2$

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Let's put everything to one side.

Subtract $r^2$ on both sides.

$x^2+a^2x^4+18ax^2+81-r^2=0$

Reorder in standard form in terms of x:

$a^2x^4+(18a+1)x^2+(81-r^2)=0$

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The discriminant is $B^2-4AC$ .

If you compare our equation to $Au^2+Bu+C$ , you should determine $A=a^2$

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$C=(81-r^2)$

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Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

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Distribute the 4a^2 to the terms in the ( ) next to it:

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$36a+1+4a^2r^2$

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We know that $r>0$ because in order it to be a circle a radius has to exist.

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We have to go further to see.

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That is,

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18a+1 is positive since a is positive so -(18a+1) is negative

2a^2 is positive (a is not 0).

So you have (negative number-positive number)/positive which is a negative since the top is negative and you are dividing by a positive.

We have confirmed are max of one solution algebraically. (It is definitely not 3 solutions.)

If r=9, then there is one solution.

If r>9, then there is two solutions as this shows:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

r=9 since our circle intersects the parabola at (0,9).

Also if (0,9) is intersection, then

$0^2+9^2=r^2$ which implies r=9.

Plugging in 9 for r we get:

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This means we should have got x=0.

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