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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

  1. A \(0.11\)
  2. B \(0.38\)
  3. C \(0.32\)
  4. D \(0.35\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.32\)

Step-by-step Solution

Detailed explanation

Given, \(P(A)=0.59, P(B)=0.30\)
and \(P(A \cap B)=0.21\)
\(\begin{aligned}
P\left(A^{\prime} \cap B^{\prime}\right) &=1-P(A \cup B) \quad[\therefore \text { De Morgan's law }] \\
=& 1-[P(A)+P(B)-P(A \cap B)] \\
=& 1-[0.59+0.30-0.21] \\
=& 1-[0.89-0.21] \\
=& 1-0.68=0.32
\end{aligned}\)