The following problem1. If X and Y's combined probability distribution is given byf(x,y)=x+y30&nbsp;for&nbsp;x=0,2,3;&nbsp;y=0,1,2&nbsp;Find&nbsp;a.&nbsp;P(X≤2,Y=1)&nbsp;b.&nbsp;P(X&gt;2,Y≤1)&nbsp;c.&nbsp;P(X&gt;Y)

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The following problem1. If X and Y&#039;s combined probability distribution is given byf(x,y)=x+y30&amp;nbsp;for&amp;nbsp;x=0,2,3;&amp;nbsp;y=0,1,2&amp;nbsp;Find&amp;nbsp;a.&amp;nbsp;P(X≤2,Y=1)&amp;nbsp;b.&amp;nbsp;P(X&amp;gt;2,Y≤1)&amp;nbsp;c.&amp;nbsp;P(X&amp;gt;Y)

saiyansruleA · Accepted Answer

Here, xyf(x,y)000200.066667300.1010.033333210.1310.133333020.066667220.133333320.166667 a) P(X≤2,Y=1)=P(X=0,Y=1)+P(X=2,Y=1) =f(0,1)+f(2,1) =0.33333+0.1 =0.133333 b) P(X&amp;gt;2,Y≤1)=P(X=3,Y=0)+P(X=3,Y=1) f(3,0)+f(3,1) =0.1+0.133333 =0.233333 c) P(X&amp;gt;Y)=P(X−Y&amp;gt;0)=P(X=2,Y=0)+P(X=3,Y=0)+P(X=2,Y=1)+P(X=3,Y=1)+P(X=3,Y=2) =0.066667+0.1+0.1+0.133333+0.166667 =0.566667

user_27qwe · Accepted Answer

Result:a. P(X≤2, Y=1) = 415.b. P(X&amp;gt;2, Y≤1) = 730.c. P(X&amp;gt;Y) = 715.Solution:a. P(X≤2, Y=1):To calculate this probability, we need to sum up the probabilities for all (X, Y) pairs that satisfy the condition X≤2 and Y=1.We have three possible pairs: (0, 1), (2, 1), and (3, 1). Therefore, we can write the probability as follows:P(X≤2, Y=1) = P((X=0, Y=1) ∪ (X=2, Y=1) ∪ (X=3, Y=1))To calculate the probability for each pair, we substitute the corresponding values into the PDF:P(X=0, Y=1) = f(0, 1) = 0+130 = 130P(X=2, Y=1) = f(2, 1) = 2+130 = 330 = 110P(X=3, Y=1) = f(3, 1) = 3+130 = 430 = 215Now, we can sum up these probabilities to get the final result:P(X≤2, Y=1) = 130 + 110 + 215 = 130 + 330 + 430 = 830 = 415.Therefore, P(X≤2, Y=1) = 415.b. P(X&amp;gt;2, Y≤1):Similar to part a, we&#039;ll sum up the probabilities for all (X, Y) pairs that satisfy the condition X&amp;gt;2 and Y≤1.The possible pairs that satisfy this condition are: (3, 0) and (3, 1).P(X&amp;gt;2, Y≤1) = P((X=3, Y=0) ∪ (X=3, Y=1))Calculating the probability for each pair:P(X=3, Y=0) = f(3, 0) = 3+030 = 330 = 110P(X=3, Y=1) = f(3, 1) = 3+130 = 430 = 215Summing up these probabilities:P(X&amp;gt;2, Y≤1) = 110 + 215 = 330 + 430 = 730.Therefore, P(X&amp;gt;2, Y≤1) = 730.c. P(X&amp;gt;Y):To calculate this probability, we need to sum up the probabilities for all (X, Y) pairs where X&amp;gt;Y.The pairs that satisfy this condition are: (2, 0), (3, 0), (3, 1), and (3, 2).P(X&amp;gt;Y) = P((X=2, Y=0) ∪ (X=3, Y=0) ∪ (X=3, Y=1) ∪ (X=3, Y=2))Calculating the probability for each pair:P(X=2, Y=0) = f(2, 0) = 2+030 = 230 = 115P(X=3, Y=0) = f(3, 0) = 3+030 = 330 = 110P(X=3, Y=1) = f(3, 1) = 3+130 = 430 = 215P(X=3, Y=2) = f(3, 2) = 3+230 = 530 = 16Summing up these probabilities:P(X&amp;gt;Y) = 115 + 110 + 215 + 16 = 230 + 330 + 430 + 530 = 1430 = 715.Therefore, P(X&amp;gt;Y) = 715.

karton · Accepted Answer

Step 1:a. To find P(X≤2,Y=1), we need to sum up the probabilities for all possible combinations of X and Y that satisfy this condition. From the given probability distribution, we have:P(X≤2,Y=1)=P(X=0,Y=1)+P(X=2,Y=1)=0+130+2+130=430=215Therefore, P(X≤2,Y=1)=215.Step 2:b. To find P(X&amp;gt;2,Y≤1), we again need to sum up the probabilities for all possible combinations that satisfy this condition:P(X&amp;gt;2,Y≤1)=P(X=3,Y=0)+P(X=3,Y=1)=3+030+3+130=730Hence, P(X&amp;gt;2,Y≤1)=730.Step 3:c. Finally, to find P(X&amp;gt;Y), we need to sum up the probabilities for all combinations where X is greater than Y:P(X&amp;gt;Y)=P(X=2,Y=0)+P(X=3,Y=0)+P(X=3,Y=1)=2+030+3+030+3+130=930=310Thus, P(X&amp;gt;Y)=310.

star233 · Accepted Answer

To solve the given problem, we need to calculate the probabilities using the given probability distribution function (PDF). Let&#039;s calculate each part step by step:a. To find P(X≤2,Y=1), we need to sum up the probabilities for all pairs (x, y) that satisfy the condition. So we have:P(X≤2,Y=1)=f(0,1)+f(2,1)+f(3,1)Substituting the values from the PDF, we get:P(X≤2,Y=1)=0+130+2+130+3+130Simplifying the expression, we have:P(X≤2,Y=1)=130+330+430=830=415Therefore, P(X≤2,Y=1)=415.b. To find P(X&amp;gt;2,Y≤1), we again need to sum up the probabilities for all pairs (x, y) that satisfy the condition. So we have:P(X&amp;gt;2,Y≤1)=f(3,0)+f(3,1)Substituting the values from the PDF, we get:P(X&amp;gt;2,Y≤1)=3+030+3+130Simplifying the expression, we have:P(X&amp;gt;2,Y≤1)=330+430=730Therefore, P(X&amp;gt;2,Y≤1)=730.c. To find P(X&amp;gt;Y), we need to sum up the probabilities for all pairs (x, y) where x is greater than y. So we have:P(X&amp;gt;Y)=f(2,0)+f(3,0)+f(3,1)+f(3,2)Substituting the values from the PDF, we get:P(X&amp;gt;Y)=2+030+3+030+3+130+3+230Simplifying the expression, we have:P(X&amp;gt;Y)=230+330+430+530=1430=715Therefore, P(X&amp;gt;Y)=715.In summary:a. P(X≤2,Y=1)=415b. P(X&amp;gt;2,Y≤1)=730c. P(X&amp;gt;Y)=715

Solve the following 1. If the joint probability distribution of X

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