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In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is $0.8$ and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is $0.95$. What is the probability of passing both?
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Let $A$ and $B$ be the events of passing $I$ and $I I$ examinations respectively.
$\therefore \quad P(A)=0.8, P(B)=0.7$
Probability of passing atleast one examination
$=1-P\left(A^{\prime} \cap B^{\prime}\right)=0.95 \quad \ldots(i)$
Now $A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}$ (De Morgan's Law)
$P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B)$
Putting this value in (i), $1-[1-P(A \cup B)]=0.95$ or $P(A \cup B)=0.95$
Further $P(A \cap B)=P(A)+P(B)-P(A \cup B)=0.8+0.7$ $-0.95=1.5-0.95=0.55$
Thus, probability that the student will pass in both the examinations $=0.55$
$\therefore \quad P(A)=0.8, P(B)=0.7$
Probability of passing atleast one examination
$=1-P\left(A^{\prime} \cap B^{\prime}\right)=0.95 \quad \ldots(i)$
Now $A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}$ (De Morgan's Law)
$P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B)$
Putting this value in (i), $1-[1-P(A \cup B)]=0.95$ or $P(A \cup B)=0.95$
Further $P(A \cap B)=P(A)+P(B)-P(A \cup B)=0.8+0.7$ $-0.95=1.5-0.95=0.55$
Thus, probability that the student will pass in both the examinations $=0.55$
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