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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  • [IIT 2008]

Fill in the blanks in following table :

$P(A)$ $P(B)$ $P(A \cap B)$ $P (A \cup B)$
$0.35$  ........... $0.25$  $0.6$

In class $XI$ of a school $40\%$ of the students study Mathematics and $30 \%$ study Biology. $10 \%$ of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

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 Only one of them will qualify the examination.