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Which one of the following propositional logic formulas is TRUE when exactly two of $p,q$ and $r$ are TRUE?

  1. $(( p  \leftrightarrow q)  \wedge  r)  \vee  (p \wedge q \wedge  \sim r)$
  2. $( \sim (p \leftrightarrow q) \wedge r)\vee (p \wedge q \wedge \sim r)$
  3. $( (p \to q) \wedge r) \vee (p \wedge q \wedge \sim r)$
  4. $(\sim (p \leftrightarrow q) \wedge r) \wedge (p \wedge q \wedge \sim r) $
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B is correct one ! Here is better clarification:

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Combination which gives False on exactly two of p,q,r are true.

Option A ---  (P,Q,R) = (F,T,T)

Option C---  (P,Q,R) = (T,F,T)

Option D---  (P,Q,R) = (T,T,F)  or (T,F,T) or (F,T,T)

Now, option B these combination give true..

Option B ---  (P,Q,R) = (T,T,F)  or (T,F,T) or (F,T,T)

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11 Answers

41 votes
41 votes
Best answer

$A.$ will be true if $P,Q,R$ are true, $((p ↔q) ∧ r)$ will return true. So "exactly two" is false
$C.$ if only $r$ is true and $p$ and $q$ are false, first part of implication itself will result in true
$D.$ if $r$ is true or false, this returns false due to $r$ and $\neg r$ present in conjunction. So, this is a CONTRADICTION. 

B is the answer. B is true if $p$ is TRUE and $q$ is FALSE or vice verse, and $r$ is true or if $p$ and $q$ are TRUE and $r$ is FALSE.

PS: Actually the question should have been "TRUE ONLY when exactly two of $p,q$ and $r$ are TRUE"

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video solution seems perfect.
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where is vid link?
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I feel the answer is right but explanation for option (A) is wrong in selected answer.

Just think what does ”q if p” sentence format say in propositional logic. It means p->q. If you see it’s truth table,it means when ‘p’ is true , ‘q’ should be compulsorily TRUE.But when ‘p’ is false , ‘q’ can assume any value.

 

→ Now here ‘p’=”EXACTLY TWO OF p,q,r are TRUE.”

→ ‘q’ = “Given option is right”.

(A) is wrong bcoz when ‘q’ and ‘r’ are TRUE and ‘p’ is false , it assumes FALSE value, which shouldn’t happen as discussed above.

(B) is right .

(C) and (D) are false for same reason as (A)

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26 votes
26 votes
the proposition logic formula is required to be true when any two of the p q r are true .so let's make truth table taking any two of p q r as true

P  Q  R

0   1   1

1    1    0

1     0    1

now write down the POS for above truth table

i.e. P'QR+PQR'+PQ'R  .

option B will satisfy the above POS expression hence answer.

4 Comments

thanks
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nice answer
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Best Answer.
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18 votes
18 votes

propositional logic formulas is TRUE when exactly two of p,q and r are TRUE means possibilities are pq or qr or pr to be TRUE 
option A) 
  ((p↔q)∧r)∨(p∧q∧∼r)
=((T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)
=((T)∧r)∨(T∧∼r)
=(r)∨(∼r)
=T 
It doesn't depend upon the value of r, so, if the value of r is T then it also to be T so the condition exactly 2 are T is false. Hence option A is FALSE.

option B)
  (∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(T↔T)∧r)∨(T∧T∧∼r)  ( take p=T and q=T) 
=(∼(T)∧r)∨(T∧∼r)
=(F∧r)∨(T∧∼r)
=(F)∨(∼r)
=∼r 
so if r=F then this is T so condition exactly 2 are T is true

 (∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(T↔q)∧T)∨(T∧q∧∼T)  ( take p=T and r=T) 
=(∼(q)∧T)∨(F)
=∼(q)∨(F)
=∼q
so if q=F then this is T so condition exactly 2 are T is true

(∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(p↔T)∧T)∨(p∧T∧∼T)  ( take q=T and r=T) 
=(∼(p)∧T)∨(F)
=∼(p)∨(F)
=∼p
so if p=F then this is T so condition exactly 2 are T is true

Hence option B is CORRECT.

option C)
   ((p→q)∧r)∨(p∧q∧∼r)
=((T→T)∧r)∨(T∧T∧∼r) ( take p=T and q=T) 
=((T)∧r)∨(T∧∼r)
=(r)∨(∼r)
=T
It doesnt depend upon value of r , so , if value of r is T then it also to be T so the condition exactly 2 are T is false . Hence option C is FALSE.

option D)
    (∼(p↔q)∧r)∧(p∧q∧∼r)
=(∼(T↔T)∧r)∧(T∧T∧∼r)  ( take q=T and r=T) 
=(∼(T)∧r)∧(T∧∼r)  
=(F∧r)∧(T∧∼r)  
=F∧(∼r)
=F  
It doesnt depend upon value of r , so , if value of r is F then it also to be F so the condition exactly 2 are T is false . Hence option D is FALSE.

3 votes
3 votes
Answer is (B) here any two propositional variables can be true and any one of p,q and r can be false .So let's start with elimination:

D is eliminated as D won't be true if p ,q will be true and r is false since there is an and between two expressions.

In C if q is false and p and r is true then the expression:(~p or q)AND r becomes false which doesn't satisfy the condition .

Coming to B it seems to be the right answer even when p is false ,q and r is true and when q is false and p and r is true.

A won't satisfy the condition if anyone p or q is false.

The key is : p<->q is actually (p->q)AND(q->p).

P->q is ~p or q.you just need to substitute T and F and check for condition.
Answer:

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