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 quailfy the examination is $0.02$. Find the probability that
(a) Both Anil and Ashima will not qualify the examination. (N)
(b) Atleast one of them will not qualify the examination. (N)
(c) Only one of them will qualify the examination. (N)

Answer 1

Let $E$ be the event that Anil will qualify the examination and $F$ be the event that Ashima will qualify the examination.

Given that:-

Probability that Anil will qualify the exam $=P\left(E\right)=0.05$

Probability that Ashima will qualify the exam $=P\left(F\right)=0.10$

Probability that both will qualify the examination $=0.02$

$\therefore P(E\cap F)=0.02$

To find:-

(a) $P\left(\text{both Anil and Ashima will not qualify the examination}\right)=?$

(b) $P\left(\text{atleast one of them will not qualify}\right)=?$

(c)$P\left(\text{only one of them will qualify}\right)=?$

Solution:-

As we know that,

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

$\Rightarrow P(E\cup F)=0.05+0.1-0.02=0.13$

• (a) $P\left(\text{both Anil and Ashima will not qualify the examination}\right)$
  

By Demorgan's law,

$P(E^{\prime }\cap F^{\prime })=P{(E\cup F)}^{\prime }=1-P(E\cup F)$

$\therefore P(E^{\prime }\cap F^{\prime })=1-0.13=0.87$

• (b) $P\left(\text{atleast one of them will not qualify}\right)=1-P(E\cap F)=1-0.02=0.98$
  
• (c) $P\left(\text{only one of them will qualify}\right)$
  

$P(E\cap F^{\prime })\cup P(E^{\prime }\cap F)=P(E\cap F^{\prime })+P(E^{\prime }\cap F)-P(E\cap F^{\prime })\cap P(E\cap F^{\prime })$

$\Rightarrow P(E\cap F^{\prime })\cup P(E^{\prime }\cap F)=P(E\cap F^{\prime })+P(E^{\prime }\cap F)[\because P(E\cap F^{\prime })\cap P(E^{\prime }\cap F)=0]$

$\Rightarrow P(E\cap F^{\prime })\cup P(E^{\prime }\cap F)=P\left(E\right)-P(E\cap F)+P\left(F\right)-P(E\cap F)$

$\Rightarrow P(E\cap F^{\prime })\cup P(E^{\prime }\cap F)=P\left(E\right)+P\left(F\right)-2P(E\cap F)$

$\Rightarrow P(E\cap F^{\prime })\cup P(E^{\prime }\cap F)=0.05+0.1-2\left(0.02\right)=0.11$