Tag: business maths

Questions Related to business maths

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:

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

  1. $\dfrac{3}{4}$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{3}{5}$

Show answer
Correct Option: D
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 $\dfrac{3}{5}.$

$A$ 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:

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

  1. $\dfrac{3}{4}$

  2. $\dfrac{2}{5}$

  3. $\dfrac{1}{4}$

  4. $\dfrac{1}{5}$

Show answer
Correct Option: B
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 $\dfrac{2}{5}.$

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?

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

  1. $0.154$

  2. $0.20$

  3. $0.254$

  4. $0.30$

Show answer
Correct Option: C
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

$\left( {p \Rightarrow q} \right) \to \left[ {\left( {r \vee p} \right) \Rightarrow \left( {r \vee q} \right)} \right]$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a contradiction

  2. a tautology

  3. a tautology and a contradiction

  4. neither a tautology nor a contradiction

Show answer
Correct Option: A

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

logical equivalence mathematical logic discrete mathematics business maths maths

  1. tautology

  2. contradiction

  3. dualoty

  4. double implication

Show answer
Correct Option: A

Which of the following is not correct ?

logical equivalence mathematical logic discrete mathematics business maths maths

  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.

Show answer
Correct Option: B,D
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.

Which of the following statement is a tautology?

logical equivalence mathematical logic discrete mathematics business maths maths

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

Show answer
Correct Option: A
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.

$ p\Rightarrow p \vee q$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.

Show answer
Correct Option: A
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

$p\Rightarrow \sim p$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.

Show answer
Correct Option: D
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

$ p\wedge  (\sim p)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.

Show answer
Correct Option: B
Explanation:

Truth table is,

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

Hence given statement is contradiction.