B is false, since by the inclusion/exclusion principle,
[tex]P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)[/tex]
By independence, we have [tex]P(X\cap Y)=P(X)P(Y)[/tex], which is zero if either of [tex]P(X)[/tex] or [tex]P(Y)[/tex] is 0, which isn't guaranteed.
Using probability of independent events, it is found that the false statement is:
[tex]P(X \cup Y) = P(X) + P(Y)[/tex]
Which is option b.
If two events, X and Y, are independent, the probability of both happening is the multiplication of the probabilities of each happening, that is:
[tex]P(X \cap Y) = P(X)P(Y)[/tex]
Additionally, the probability of at least one happening is:
[tex]P(X \cup Y) = P(X) + P(Y) + P(X \cap Y)[/tex]
Since [tex]P(X \cap Y) = P(X)P(Y)[/tex]
[tex]P(X \cup Y) = P(X) + P(Y) + P(X)P(Y)[/tex]
Which means that statement b is false.
A similar problem is given at https://brainly.com/question/24174994
