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Home  >>  CBSE XI  >>  Math  >>  Probability

A and B are events such that $P(A)=0.42,P(B)=0.48,$P(A and B)=0.16.Determine P(A or B)

$\begin{array}{1 1}(A)\;0.74\\(B)\;0.64\\(C)\;0.54\\(D)\;\text{None of these}\end{array} $

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1 Answer

Toolbox:
  • The formula used to solve this problem $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Step 1:
P(A)=0.42
P(B)=0.48
P(A and B)=0.16
Step 2:
P(A or B)=$P(A \cup B)$
$\qquad\quad\;=P(A)+P(B)-P(A\cap B)$
$\qquad\quad\;=0.42+0.48-0.16$
$\qquad\quad\;=0.90-0.16$
$\qquad\quad\;=0.74$
Hence (A) is the correct answer.
answered Jul 1, 2014 by sreemathi.v
 

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