Question

Two events \(A\) and \(B\) are said to be independent, if:

• A. \(A\) and \(B\) are mutually exclusive.
• B. \(P(A'B')=[1–P(A)] [1–P(B)]\)
• C. \(P(A)=P(B)\)
• D. \(P(A)+P(B)=1\)

Answer 1

The Correct Option is B

The correct answer is B:\(P(A'B')=[1–P(A)] [1–P(B)]\)
Two events A and B are said to be independent, if \(P(AB) = P(A) × P(B)\)
Consider the result given in alternative B.
\(P(A'B')=[1–P(A)] [1–P(B)]\)
\(\implies P(A'\cap B')=1-P(A)-P(B)+P(A).P(B)\)
\(\implies 1-P(A\cup B)=1-P(A)-P(B)+P(A).P(B)\)
\(\implies P(A\cup B)=P(A)+P(B)-P(A).P(B)\)
\(\implies P(A)+P(B)-P(AB)=P(A)+P(B)-P(A).P(B)\)
\(\implies P(AB)=P(A).P(B)\)
This implies that A and B are independent, if \(P(A'B')=[1–P(A)] [1–P(B)]\)
Distracter Rationale
A. Let \(P (A) = m, P (B) = n, 0 < m, n < 1\)
\(A\) and \(B\) are mutually exclusive
\(\therefore A\cap B=\phi\)
\(\implies P(AB)=0\)
However, \(P(A).P(B)=mn\neq0\)
\(\therefore P(A).P(B) \neq P(AB)\)
C. Let \(A\): Event of getting an odd number on throw of a die = {1, 3, 5}
\(\implies P(A)=\frac{3}{6}=\frac{1}{2}\)
B: Event of getting an even number on throw of a die = {2, 4, 6}
\(\implies P(B)=\frac{3}{6}=\frac{1}{2}\)
\(\)Here, \(A\cap B=\phi\)
\(\therefore P(AB)=0\)
\(P(A).P(B)=\frac{1}{4} \neq0\)
\(\implies P(A).P(B) \neq P(AB)\)
D. From the above example, it can be seen that,
\(P(A)+P(B)=\frac{1}{2}+\frac{1}{2}\)
However, it cannot be inferred that \(A\) and \(B\) are independent.
Thus, the correct answer is B.