Answer:
0.3125
Step-by-step explanation:
From the full question in your book, the probability can be computed as:
[tex]P(X+Y=6) = P \{X=1,Y=5\} \cup \{X=2,Y=4\} \cup \{X=3,Y=3\} \cup \{X=4,Y=2\} \cup \{X=5,Y=1\}[/tex]
[tex]= P(X=1,Y=5) + P(X=2,Y=4)+P(X=3,Y=3)+P(X=4,Y=2)+P(X=5,Y=1)[/tex]
since all events are enclosed in brackets, they are mutually exclusives.
Then:
[tex]= P(X=1)P(Y=5)+P(X=2)P(Y=4)+P(X=3)P(Y=3)+P(X=4)P(Y=2)+P(X=5)P(Y=1)[/tex]
as X and Y are independent;
= [tex](0.25\times0.25 + 0.25\times0.25 + 0.25\times0.25+0.25\times0.25 + 0.25\times0.25)[/tex]
= 0.3125
Therefore;
P(X+Y= 6) = 0.3125
