Answer:
a) Joint ptobability distribution
![\begin{pmatrix} &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7D%20%20%26Y%3D0%26Y%3D1%26Y%3D2%26Y%3D3%26Y%3D4%5C%5CX%3D0%260.3%260.05%260.025%260.025%260.1%5C%5C%20X%3D1%260.18%260.03%260.015%260.015%260.06%20%5C%5C%20X%3D2%260.12%260.02%260.01%260.01%260.04%5Cend%7Bpmatrix%7D)
b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56
c) P(X + Y = 0)=0.3
d) P(X + Y <= 1)=0.53
Step-by-step explanation:
We have to construct the joint probability table with the marginal probabilities of X and Y.
X can take values from 0 to 2, and Y can take values from 0 to 4.
We can calculate each point of the joint probability as:
![P(x,y)=P_x(x)*P_y(y)](https://tex.z-dn.net/?f=P%28x%2Cy%29%3DP_x%28x%29%2AP_y%28y%29)
Then, the joint probabilities are:
X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3
X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05
X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025
X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025
X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1
X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18
X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03
X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015
X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015
X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06
X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12
X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02
X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01
X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01
X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04
We can write it in the form of a matrix:
![\begin{pmatrix} &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7D%20%20%26Y%3D0%26Y%3D1%26Y%3D2%26Y%3D3%26Y%3D4%5C%5CX%3D0%260.3%260.05%260.025%260.025%260.1%5C%5C%20X%3D1%260.18%260.03%260.015%260.015%260.06%20%5C%5C%20X%3D2%260.12%260.02%260.01%260.01%260.04%5Cend%7Bpmatrix%7D)
b) From the joint probability P(X<= 1 and Y <= 1) is equal to
![P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56](https://tex.z-dn.net/?f=P%28X%5Cleq%201%20%5C%26%20Y%20%5Cleq%201%29%3DP%280%2C0%29%2BP%280%2C1%29%2BP%281%2C0%29%2BP%281%2C1%29%5C%5C%5C%5CP%28X%5Cleq%201%20%5C%26%20Y%20%5Cleq%201%29%3D0.3%2B0.05%2B0.18%2B0.03%3D0.56)
We can calculate P(X<= 1) * P(Y<=1)
![P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56](https://tex.z-dn.net/?f=P_x%28X%5Cleq%201%29%3DP_x%280%29%2BP_x%281%29%3D0.5%2B0.3%3D0.8%5C%5C%5C%5CP_y%28Y%5Cleq1%29%3DP_y%280%29%2BP_y%281%29%3D0.6%2B0.1%3D0.7%5C%5C%5C%5CP_x%28X%5Cleq1%29%2AP_y%28Y%5Cleq1%29%3D0.8%2A0.7%3D0.56)
Both calculations give the same result.
c) Probability of no violations
![P(X+Y=0)=P(0,0)=0.3](https://tex.z-dn.net/?f=P%28X%2BY%3D0%29%3DP%280%2C0%29%3D0.3)
d) P(X + Y <= 1)
![P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53](https://tex.z-dn.net/?f=P%28X%2BY%20%5Cleq%201%29%3DP%280%2C0%29%2BP%280%2C1%29%2BP%281%2C0%29%5C%5C%5C%5CP%28X%2BY%20%5Cleq%201%29%3D0.3%2B0.05%2B0.18%3D0.53)