KCET · Maths · Probability
If \(P(A)=0.59, P(B)=0.30\) and \(P(A \cap B)=0.21\) then \(P\left(A^{\prime} \cap B^{\prime}\right)\) is equal to
- A \(0.11\)
- B \(0.38\)
- C \(0.32\)
- D \(0.35\)
Answer & Solution
Correct Answer
(C) \(0.32\)
Step-by-step Solution
Detailed explanation
Given, \(P(A)=0.59, P(B)=0.30\)
and \(P(A \cap B)=0.21\)
\(\begin{aligned}
P\left(A^{\prime} \cap B^{\prime}\right) &=1-P(A \cup B) \quad[\therefore \text { De Morgan's law }] \\
=& 1-[P(A)+P(B)-P(A \cap B)] \\
=& 1-[0.59+0.30-0.21] \\
=& 1-[0.89-0.21] \\
=& 1-0.68=0.32
\end{aligned}\)
and \(P(A \cap B)=0.21\)
\(\begin{aligned}
P\left(A^{\prime} \cap B^{\prime}\right) &=1-P(A \cup B) \quad[\therefore \text { De Morgan's law }] \\
=& 1-[P(A)+P(B)-P(A \cap B)] \\
=& 1-[0.59+0.30-0.21] \\
=& 1-[0.89-0.21] \\
=& 1-0.68=0.32
\end{aligned}\)
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