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Questions Related to business maths

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ for post No. three. The probability of A's being selected for at least one post is:

  1. $\dfrac{3}{4}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{3}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A)=\dfrac{1}{4}$    $P(B)=\dfrac{1}{3}$     $P(C)=\dfrac{1}{5}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$

                                 $=\dfrac{15+20+12-5-3-4+1}{60}$
                                 $=\dfrac{3}{5}.$
Hence, the answer is $\dfrac{3}{5}.$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

$A$ is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ for post No. three. the probability of $A$'s being selected for none of the posts is:

  1. $\dfrac{3}{4}$
  2. $\dfrac{2}{5}$
  3. $\dfrac{1}{4}$
  4. $\dfrac{1}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$
                                 $=\dfrac{3}{5}.$
$\Rightarrow P$ ( not getting selected for none of the post ) $=1-\dfrac{3}{5}$
                                                                                   $=\dfrac{2}{5}$
Hence, the answer is $\dfrac{2}{5}.$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:

Sum  Frequency
2 14
3 30
4 43
5 55
6 72
7 75
8 70
9 53
10 46
11 28
12 15

If the dice are thrown once more, what is the probability of getting a sum between 8 and 12?

  1. $0.154$
  2. $0.20$
  3. $0.254$
  4. $0.30$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If we rolled dice around 500 times then it can be considered as its expected outcome.
Hence,
Probability of obtaining between 8 and 12 as sum = 53 + 46 +28/500= 0.254

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$(p \wedge \sim q)\wedge (\sim p \vee q)$ is

  1. tautology

  2. contradiction

  3. dualoty

  4. double implication
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Evaluating the expression using truth tables or logical equivalences shows that it evaluates to false for all truth values of the variables. The components simplify such that a proposition and its negation are strictly contradicted, yielding a contradiction.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following is not correct ?

  1. $p \vee \sim p $ is a tautology.
  2. $\sim (\sim p) \leftrightarrow p$ is a tautology.
  3. $p \wedge \sim p $ is a contradiction.
  4. $([(p \wedge p ) \rightarrow q] \rightarrow p)$ is a tautology.
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

$\because [(p \wedge p ) \rightarrow q ] \rightarrow p \equiv ( p

\rightarrow q ) \rightarrow p   (\because p \wedge p \equiv p)$

when $p$ is false and $q$ is true (or false) then

$(p \rightarrow q) $ is true i.e. $(p \rightarrow q ) ]\rightarrow p$ is false

Hence $[(p \wedge p ) \rightarrow q] \rightarrow p$ is not a tautology.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following statement is a tautology?

  1. $(\sim p \vee \sim q ) \vee ( p \vee \sim q )$
  2. $(\sim p \vee \sim q ) \wedge (p \vee \sim q )$
  3. $\sim p \wedge (\sim p \vee \sim q )$
  4. $\sim q \wedge (\sim p \vee \sim q )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\because (\sim p  \vee \sim q) \vee (p  \vee \sim q) $

$\equiv (\sim p \vee p ) \vee (p \vee \sim q )$    (by distributive law)

$ \equiv t \vee \sim q \equiv t $                      t is a tautology

Hence $(\sim p  \vee \sim, q ) \vee (p  \vee \sim q )$ is a tautology.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$ p\Rightarrow p \vee q$ is

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$p$       $q$      $p\vee q$       $p\rightarrow (p\vee q)$
T T     T            T
T F    T            T
F T    T              T
F F     F            T

Since, all the entries in the last column has true value. So, the given statement is a tautology

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$p\Rightarrow \sim p$ is

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
 $p$  $\sim p$  $p\Rightarrow \sim p$
 T  F  F
 F  T  T

So the result of the Truth table show that $p\Rightarrow \sim p$ is neither a tautology not a contradiction

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$ p\wedge  (\sim p)$ is

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Truth table is,

$p$         $\sim p$ $p\wedge (\sim p)$
T                      F                      F
F T F

Hence given statement is contradiction.