Question

The joint probability distribution of x and y is given below:

x\rightarrow y\downarrow 0 1 2
0 \frac{1}{12} \frac{1}{6} \frac{1}{24}
1 \frac{1}{4} \frac{1}{4} \frac{1}{40}
2 \frac{1}{8} \frac{1}{20} __
3 \frac{1}{120} __ \frac{1}{120}

Find (i) )P(x=1,=2)
 (ii) )P(x=0,1\leq y<3)
 (iii) )P(x+y\leq1)
 (iv) )P(x>y)

12 Mar 2024
Answer :
Word Count : 226
Let's first reconstruct the joint probability table with the missing values: | \( x \) ↓ | \( y \) → | 0 | 1 | 2 | |------------|------------|------------|------------| | 0 | \( \frac{1}{12} \) | \( \frac{1}{6} \) | \( \frac{1}{24} \) | | 1 | \( \frac{1}{4} \) | \( \frac{1}{4} \) | \( _________ __________ __________ __________ ____ ___ ____ _______ ___ ____ ___.
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