Q. If the probability for $A$ to fail in an examination is 0.2 and that for $B$ is 0.3, then the probability that either $A\ or\ B$ fails, is
A) 0.38
B) 0.44
C) 0.50
D) 0.94
Explanation
(b)
$P\left( \overline{A} \right) = 0.2,\ P\left( \overline{B} \right) = 0.3$
$P(A) = 0.8,\ P(B) = 0.7$
$\therefore$ Required probability
$= P\left( \overline{A} \right)P(B) + \ P(A)\ P\left( \overline{B} \right) + P\left( \overline{A} \right)P\left( \overline{B} \right)$
$= 0.2 \times 0.7 + 0.8 \times 0.3 + 0.2 \times 0.3$
$= 0.44$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.