Question

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 $0.02$. Then, which of the following is correct?
I. Probability that Anil and Ashima will not qualify the examination is $0.87$.
II. Probability that atleast one of them will not qualify the examination is $0.98$.
III. Probability that only one of them will qualify the examination is $0.11$.

• A. Only I is correct
• B. Only II is correct
• C. Only III is correct
• D. All are correct

Answer 1

Answer: (D)

Let $E$ and $F$ denote the events that Anil and Ashima will qualify the examination, respectively.
Given that,
$P(E)=0.05, P(F)=0.10 \text { and } P(E \cap F)=0.02 \text {. }$
Then,
I. 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 $E$ ', i.e., Anil will not qualify the examination and $F$ ' is 'not $F$ ', 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, $P\left(E^{\prime} \cap F^{\prime}\right)=P(E \cup F)=1-P(E \cup F)$
$=1-0.13=0.87$
II. $P$ (atleast one of them will not qualify)
$=1-P$ (both of them will qualify)
$=1-0.02=0.98$
III. The event only one of them will qualify the examination is same as the event either (Anil will qualify and Ashima will not qualify) or (Anil will not qualify and Ashima will qualify) i.e., $E \cap F^{\prime}$ or $E^{\prime} \cap F$, where $E \cap F^{\prime}$ and $E^{\prime} \cap F$ are mutually exclusive.
Therefore, $P$ (only one of them will qualify)
$=P\left(E \cap F^{\prime}\right.$ or $\left.E^{\prime} \cap F\right)$
$=P\left(E \cap F^{\prime}\right)+P\left(E^{\prime} \cap F\right)$
$=P(E)-P(E \cap F)+P(F)-P(E \cap F)$
$=0.05-0.02+0.10-0.02=0.11$