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Questions Related to business maths

Which of the following is a tautology?

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  1. $p\implies p\wedge q$

  2. $p\implies p\vee q$

  3. $(p\vee q)\implies(p\wedge q)$

  4. None of these

Show answer
Correct Option: B
Explanation:

A tautology is a statement that is always true.
The statement $ p\Longrightarrow p\vee q $ is read as " if p is true, then either p or q is true. " as the symbol $ \vee  $ denotes OR.
Hence, the given statement is a tautology.

Which of the following statements is/are true?

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  1. $p\wedge (\sim p)$ is a contradiction.

  2. $(p\rightarrow q)\Leftrightarrow (\sim q \rightarrow \sim p)$ is a contradiction.

  3. $\sim(\sim p) \Leftrightarrow p$ is a tautology.

  4. $p\vee (\sim p)$ is a tautology.

Show answer
Correct Option: A,C,D
Explanation:

A,C,D obvious and for B
p $\rightarrow$ q is same as $\sim$ q $\rightarrow$ $\sim$p
$\therefore$ it is tautology not contradiction.

If $p$ is any statement $t$ and $c$ are tautology and contradiction respectively, then which of the following is(are) correct?

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  1. $p\wedge t \equiv p$

  2. $p \wedge c \equiv c$

  3. $p\vee t \equiv c$

  4. $p \vee c \equiv p$

Show answer
Correct Option: A,B,D
Explanation:

Tautology is the preposition which is always true and contradiction is a preposition which is always false.
Here $\wedge \equiv AND, \vee \equiv OR$, $T$=true and $F$=false
Given statement=$p$, tautology=$t$ and contradiction=$c$
a)$p\wedge t\equiv p$ irrespective of value of $p$ as $t=T$ always
And if $p=T$ then $T\wedge T=T$ and if $p=F$ then $F\wedge T=F$
Thus option (a) is correct.
b)$p\wedge c\equiv c$ irrespective of value of $p$ as$ c=F$ always
And if $p=T$ then $T\wedge F=F$ and if $p=F$ then $F\wedge F=F$
Thus option (b) is correct
c)$p\vee t\equiv c$
If $p=T$ then $T\vee T\equiv T\neq c$ and if $p=F$ then $F\vee T\equiv T \neq c$
Thus option (c) is not correct
d)$p\vee c\equiv p$
If $p=T$ then $T\vee F\equiv T$ and if $p=F$ then $F\vee F\equiv F$
Thus it depends on the value of $p$
Hence option (d) is correct

Which one of the following statements is a tautology?

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  1. $\left( p\vee q \right) \rightarrow q$

  2. $p\vee (p\rightarrow q)$

  3. $ p\vee (q\rightarrow p)$

  4. $p\rightarrow (p\rightarrow q)$

Show answer
Correct Option: B
Explanation:
$ p$  $q$  $p\to q$  $q\to p$ $pvq$  $p\to (p\to q)$  $(pvq)\to q$  $pv(p\to q)$   $pv(q\to q$
 T  T  T  T  T  T  T  T  T
 T  F  F  T  T  F  F  T  T
 F  T  T  F  T  T  T  T  F
 F  F  T  T  F  T  T  T  T

$\left( p\vee q \right) \rightarrow q$

If $p$ and $q$ are two statement, then $(p  \wedge \sim q) \wedge   (\sim p  \wedge   q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a fallacy

  2. a tautology

  3. neither tautology nor a fallacy

  4. none of these

Show answer
Correct Option: A
Explanation:
$p$         $q$          $p \land \sim q$       $q \land \sim p $          $(p \land \sim q ) \land(q \land \sim p )$
T T     F       F          F
T F     T       F          F
F T      F       T          F
F F     F       F          F


$\therefore\ (p \land \sim q ) \land(q \land \sim p )$ is a fallacy

Negation of ''A is in Class $X^{th}$ or B is in $XII^{th}$'' is

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  1. A is not in class $X^{th}$ but B is in $XII^{th}$.

  2. A is not in class $X^{th}$ but B is not in $XII^{th}$.

  3. Either A is not in class $X^{th}$ or B is not in $XII^{th}$.

  4. none of these

Show answer
Correct Option: B
Explanation:

$\sim (p \lor q)=\sim p \land \sim q$
negaion of "A is in $X^{th}$ and B is in $XII^{th}$" is 
"A is not in $X^{th}$ and B is not in $XII^{th}$

Box I contains $2$ white and $3$ red balls and box II contains $4$ white and $5$ red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box II, is?

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  1. $\cfrac{54}{44}$

  2. $\cfrac{54}{14}$

  3. $\cfrac{54}{104}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Probability that the ball drawn is red and from ! =$P(R/A)$

$P(R/A) = \cfrac{P(A/R)\times P(A)}{P(B/R)\times P(B) + P(A/R) \times P(A)}$
$P(R/A) = \cfrac{3}{5}, P(R/B) = \cfrac{5}{9}$
$P(A) = \cfrac{1}{2}, P(B) = \cfrac{1}{2}$
$P(R/A) = \cfrac{\cfrac{3}{5}\times \cfrac{1}{2}}{\cfrac{5}{9}\times \cfrac{1}{2} + \cfrac{3}{5} \times \cfrac{1}{2}} = \cfrac{54}{104}$

An arrangement is selected at random from all possible arrangements of five digits written from the digits $0,1,2,3,\cdots 9$ with repetition. The probability that the randomly selected arrangement will have largest number $'8'$ given that the smallest number is $'4'$ is :

business maths probability - iii baye's theorem bayes theorem probability and probability distribution

  1. $\dfrac {1253}{6480}$

  2. $\dfrac {513}{4651}$

  3. $\dfrac {2881}{6480}$

  4. $\dfrac {1320}{4651}$

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Correct Option: C

If two events A and B are such that $P(A')=0.3, P(B)=0.5$ and $P(A\cap B)=0.3$ then $P(B/A\cup B)$=

business maths probability - iii baye's theorem bayes theorem probability and probability distribution

  1. 0.375

  2. 0.32

  3. 0.31

  4. 0.28

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Correct Option: A

The number of committees formed by taking $5men$ and $5women$ from $6women$ and $7men$ are

business maths probability - iii baye's theorem bayes theorem probability and probability distribution

  1. 252

  2. 125

  3. 126

  4. 64

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Correct Option: C