1). Since it is not clear how the joint probability function looks like, consider two cases:
a). f(x,y)=5(2x+3y)2. The following condition must be satisfied: ∬R2f(x,y)dydx=1. We receive: 52∫01∫012x+3y1dydx=52∫0131ln(2x+3y)∣01dx=152∫01(ln(2x+3)−ln(2x))dy. Consider separately integral: ∫ln(t)dt=tln(t)−∫dt=tln(t)−t+C. Using the latter, we get: 52∫01∫012x+3y1dydx=152∫01(ln(2x+3)−ln(2x))dx=152(21(2x+3)ln(2x+3)−23−xln(2x))∣01≈0.224
Thus, it is not a valid joint probability function.
b). Consider the function: f(x,y)=52(2x+3y). We will get: 52∫01∫01(2x+3y)dydx=52∫01(2xy+23y2)∣01dx=52∫01(2x+23)dx=52(x2+23x)∣01=1. Thus, it is a valid joint probability function.
Remind the formula: P(A∣B)=P(B)P(A∩B). From the latter we get: P(31<X<32∣Y>43)=P(Y>43)P(31<X<32,Y>43).
Consider separately each probability in the last formula: P(31<X<32,Y>43)=∫3132∫43152(2x+3y)dydx=52∫3132(2xy+23y2)431dx=52∫3132(2x+23−23x−3227)dx=52∫3132(21x+3221)dx=52(41x2+3221x)3132=24029
P(Y>43)=52∫01∫431(2x+3y)dydx=52∫01(2xy+23y2)∣431dx=52∫01(2x+23−23x−3227)dx=52∫01(21x+3221)dx=52(41x2+3221x)∣01=52(41+3221)=8029
Thus, the probability is the following: P(31<X<32∣Y>43)=24029⋅2980=31.
Answer: P(31<X<32,Y>43)=31
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