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14.Probability
easy
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P (A$ or $B).$
A
$0.74$
B
$0.74$
C
$0.74$
D
$0.74$
Solution
It is given that $P ( A )=0.42$, $P ( B )=0.48$, $P ( A $ and $B )=0.16$
We know that $P ( A$ or $B )= P ( A )+ P ( B )- P ( A $ and $B )$
$P ( A$ or $B )=0.42+0.48-0.16=0.74$
Standard 11
Mathematics