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3 years ago

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X and Y are loss random variables, with X discrete and Y continuous. The joint density function of X and Y is f(x, y) = [(x+1)e-

y/2] / 12 for x = 0, 1, 2 and 0 < y < [infinity]. Find the probability that the total loss, X + Y is less than 2.

1 answer:

STALIN [3.7K]3 years ago

3 0

Answer:

The probability that the total loss, X + Y is less than 2 is P=0.235

Step-by-step explanation:

We know the joint density function:

$f(x,y)=\frac{(x+1)e^{-y/2}}{12}$

To find the probability that (X+Y)<2, we can divide this in two steps.

- When X=0, Y should be less than 2. This is P(X=0,Y<2).

- When X=1, Y should be less than 1. This is P(X=1, Y<1).

We can calculate P(X=0,Y<2) as:

$P(X=0,Y$

We can calculate P(X=1,Y<1) as:

$P(X=1,Y$

Then

$P(X+Y$

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I'll assume the ODE is

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take the ansatz

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The answer is 98 small and 57 large cups.

s - the number of small cups
l - the number of large cups

<span>Ashley sold a total of 155 cups: s + l = 155
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</span>s + l = 155
1.25 * s + 2.50 * l = 265
________
s = 155 - l
1.25 * s + 2.50 * l = 265
________
1.25 * (155 - l) + 2.50 * l = 265
193.75 - 1.25 * l + 2.50 * l = 265
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______
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$If\ g\ is\ proportional\ to\ the\ t^2\ we\ have\ proportion\\\\
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