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2. (10 points) Let X and Y have the joint pdf \begin{equation*} f(x,y) = \frac{3}{2}x^2(1-|y|), \quad -1 < x < 1, \quad -1 < y < 1. \end{equation*} • Find $P((x,y) \in A)$, $A = \{(x,y): 0 < x < 1, 0 < y < 1\}$ 3. (10 points) Each of the 7 students in a class is given a fair 7-sided die. In addition, each student is numbered from 1 to 7. (a) If the students roll their dice, what is the probability that there is at least one \"match\" (e.g., student 4 rolls at 4)? (b) If you are a member of this class, what is the probability that at least one of the other 6 students rolls the same number as you do? 

          2. (10 points) Let X and Y have the joint pdf
\begin{equation*}
f(x,y) = \frac{3}{2}x^2(1-|y|), \quad -1 < x < 1, \quad -1 < y < 1.
\end{equation*}
• Find $P((x,y) \in A)$, $A = \{(x,y): 0 < x < 1, 0 < y < 1\}$
3. (10 points) Each of the 7 students in a class is given a fair 7-sided die. In addition, each student is
numbered from 1 to 7.
(a) If the students roll their dice, what is the probability that there is at least one \"match\" (e.g., student
4 rolls at 4)?
(b) If you are a member of this class, what is the probability that at least one of the other 6 students rolls
the same number as you do?
Show more…
2. (10 points) Let X and Y have the joint pdf f(x,y) = (3)/(2)x^2(1-|y|), -1 < x < 1, -1 < y < 1. • Find P((x,y) ∈ A), A = {(x,y): 0 < x < 1, 0 < y < 1} 3. (10 points) Each of the 7 students in a class is given a fair 7-sided die. In addition, each student is numbered from 1 to 7. (a) If the students roll their dice, what is the probability that there is at least one m̈atch(̈e.g., student 4 rolls at 4)? (b) If you are a member of this class, what is the probability that at least one of the other 6 students rolls the same number as you do?

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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2.10 points Let X and Y have the joint pdf f2,=1-|y-1<x<1,-1<y<1. FindPxyCAA={x0<x<1,0<<1} 3.10 points Each of the 7 students in a class is given a fair 7-sided die.In addition each student is numbered from1to 7. a If the students roll their dice,what is the probability that there is at least one matcheg.student 4rolls at4)? (b If you are a member of this class,what is the probability that at least one of the other 6 students rolls the same number as you do?
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Transcript

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00:01 So in this question we have that x is the number shown on a red fair dye, and y is equal to zero if the two dice share the same number.
00:26 It's equal to one if the number on the green die is bigger than the number on the red die.
00:34 But the number on the green die, the number on the red dye is x.
00:38 And it's two if the number on the red dye is greater than the number on the green die.
00:45 So what we can say, so let's write down a joint probability distribution.
00:52 So x can be 1, 2, 3, 4, 5 or 6, and y can be 0, 1 or 2.
01:10 So let's draw out this table.
01:16 Now let's fill it in.
01:17 So if x is equal to 1, well the probability y is 0 is always going to be the same, because the probability that the two of them are equal is always going to be 1 6th given each value of x.
01:32 So that means that multiplying that by a 6th, we get 1 over 36 for each, that y will be 0, and that x will show any particular number.
01:43 So this is 1 over 36, 1 over 36.
01:49 Etc.
01:55 Now for y to be one, that means the number on the green dye has to be bigger than x.
02:00 So for x equals 1, there are five numbers that are bigger than x.
02:06 So, and then the probability that x is each of these numbers is going to be one -sixth.
02:10 So we get five -six times one -sixth.
02:12 That gives us five over 36.
02:15 Here we have four six, because there's four numbers that are bigger than x, and we have a one -sixth chance of x being two.
02:21 So we have four over 36, three, three, over 36, 2 over 36, 1 over 36, and then 0, because if x is 6, there's no way that the green dye can be bigger than that.
02:35 But then if x is bigger than the green die, well, if x is 1, that can't happen.
02:40 If x is 2, then we've got a 1 over 36 chance of that being of this situation occurring.
02:46 And this is just going to look like the mirror image of the line above.
02:54 And now let's see whether these all add up to 1, because that will tell us, give us a check of whether this all works out...
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