GATE CSE 2003 | Question: 3

Question

Let $P(E)$ denote the probability of the event $E$. Given $P(A) = 1$, $P(B) =\dfrac{1}{2}$, the values of $P(A\mid B)$ and $P(B\mid A)$ respectively are

• A. $\left(\dfrac{1}{4}\right),\left(\dfrac{1}{2}\right)$
• B. $\left(\dfrac{1}{2}\right),\left(\dfrac{1}{4}\right)$
• C. $\left(\dfrac{1}{2}\right),{1}$
• D.   ${1},\left(\dfrac{1}{2}\right)$

Answer 1

Consider an example :

In a class there are 100 strudents(50 girls and 50 boys)
the teacher selects a monitor for the class.
A=event that a student is choosen
P(A)=1

B=event that a boy is choosen
P(B)=1/2.
now,P(A/B)=the probabiilty that a student is chosen given a boy is choosen is always 1.

P(B/A)=the probabiilty that a boy is chosen given a student  is choosen is always 1/2.(As it can be a boy or a girl).

Answer is D

Answer 2

P(A)=1,means A is the set of all possible outcomes. P(B)=½  and A∩B=B so P(A∩B)=P(B)=½

now P(A|B)=P(A∩B)/P(B)=(½ )/(½ )=1 and P(B|A)=(½ )/1 =½

So Option D is correct