Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10 .$ The probability that both will qualify the examination is $0.02 .$ Find the probability that Both Anil and Ashima will not qualify the examination.

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Let $E$ and $F$ denote the events that Anil and Ashima will qualify the examination, respectively. Given that

$P(E)=0.05$,  $P(F)=0.10$ and $P(E \cap F)=0.02$

Then

The event ' both Anil and Ashima will not qualify the examination' may be expressed as $E ^{\prime} \cap F^{\prime}$

since, $E ^{\prime}$ is 'not $E^{\prime},$ i.e., Anil will not qualify the examination and $F ^{\prime}$ is 'not $F^{\prime}$, i.e. Ashima will not qualify the examination.

Also $E ^{\prime} \cap F ^{\prime}=( E \cup F )^{\prime}$     (by Demorgan's Law)

Now $P ( E \cup F )= P ( E )+ P ( F )- P ( E \cap F )$

or   $P(E \cup F)=0.05+0.10-0.02=0.13$

Therefore $P\left(E^{\prime} \cap F^{\prime}\right)$ $=P(E \cup F)^{\prime}$ $=1-P(E \cup F)=1-0.13=0.87$

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Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$

Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$

Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$

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