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Questions Related to business maths

A & B are sharp shooters whose probabilities of hitting a target are $\displaystyle \frac{9}{10}$ & $\displaystyle \frac{14}{15}$ respectively. If it is knownthat exactly one of them has hit the target, then the probability that it was hit by A is equal to

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

  1. $\displaystyle \frac{24}{55}$

  2. $\displaystyle \frac{27}{55}$

  3. $\displaystyle \frac{9}{23}$

  4. $\displaystyle \frac{10}{23}$

Show answer
Correct Option: C
Explanation:

$E _1$ : only A hits the target
$E _2$ : only B hits the target
$E$ : exactly one hits the target.
$\therefore \displaystyle P(E _1 / E) = \frac{P(E _1). P (E / E _1)}{P (E _1). P (E/ E _1) + P (E _2). P (E/ E _2)}$
$=

\displaystyle \frac{\displaystyle \frac{9}{10} \times \frac{1}{15}}{

\displaystyle \frac{9}{10} \times \frac{1}{15} + \frac{14}{15} \times

\frac{1}{10}}\ = \dfrac{9}{23}$

A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8. from house B, 5 from  house C, 2 from house 0 and rest from house E. A single student is selected at random ,to be the class monitor. The probability that the selected student is not from A, Band C is?

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

  1. $\displaystyle \frac{4}{23}$

  2. $\displaystyle \frac{6}{23}$

  3. $\displaystyle \frac{8}{23}$

  4. $\displaystyle \frac{17}{23}$

Show answer
Correct Option: B
Explanation:
Total number of students, n(S) = 23

Number of students in houses A,B and C 

                                = 4+8+5 = 17 

∴ Remaining  students = 23 - 17 = 6 n(E) = 6

So, probability that the selected students is not from A,B and C

$P(E)=\dfrac{6}{23}$

A man is know to speak the truth $3$ out if $4$ times. He throws a die and reports that it is a six. The probability that it is actually a six is:

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

  1. $\dfrac{3}{8}$

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

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

  4. None of these

Show answer
Correct Option: A
Explanation:
Let E be the event that the man reports that six occurs in the throwing of the die and let $S _1$ be the event that six occurs and $S _2$ be the event that six does not occur.
$P(S _1)=\dfrac 16, P(S _2)= 1-\dfrac 16=\dfrac 56$
$P(E/S _1)$=probability that the man reports that six occurs when 6 has actually occurred on the die.
$P(E/S _1)$=probability that the man speaks the truth=$\dfrac 34$
$P(E/S _2)$=probability that the man reports that six occurs when 6 has not actually occurred on the die.
$P(E/S _2)$=probability that the man does not speak the truth
$\implies = 1−\dfrac 34=\dfrac 14$
Hence by Baye's theorem, we get,
$P(S _1/E)$=Probability that the report of the man that six has occurred is actually a six.
$P(S _1/E)=\dfrac {P(S _1).P(E/S _1)}{P(S _1)P(E/S _1)+P(S _2).P(E/S _2)}\\\implies = \dfrac {\dfrac 16\times \dfrac 34}{\dfrac 16\times \dfrac 34+\dfrac 56\times \dfrac 14}=\dfrac 18\times \dfrac {24}{8}=\dfrac {3}{8}$

If $P(A)=0.40,P(B)=0.35$ and $P\left( A\cup B \right) =0.55$, then $P(A/B)=$ ____

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

  1. $\cfrac { 1 }{ 5 } $

  2. $\cfrac { 8 }{ 11 } $

  3. $\cfrac { 4 }{ 7 } $

  4. $\cfrac { 3 }{ 4 } $

Show answer
Correct Option: C
Explanation:
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$P(A\cup B)=0.55$
$P(A)=0.40$
$P(B)=0.35$
$0.55=0.40+0.35-P(A\cap B)$
$P(A\cap B)=0.4+0.35-0.55=0.2$
Now
$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$
$P(A|B)=\dfrac{0.2}{0.35}$
$P(A|B)=\dfrac{4}{7}$

There are $n$ distinct white and $n$ distinct black balls. The number of ways of arranging them in a row so that neighbouring balls are of different colours is:

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

  1. $n+1C _{2}$

  2. $(2)(n\ !)^{2}$

  3. $\dfrac{(n\ !)}{2}$

  4. $none\ of\ these$

Show answer
Correct Option: B
Explanation:

Possible arrangements are BWBW....... or WBWB......

For combination BWBW..... , n blacks can permutate in n! ways and n white balls can permutate in n! ways
Total number of arrangements are (n!)(n!)
Since there are two possible arrangements total ways =$2{ (n!) }^{ 2 }$

An artillery target may be either at point $I$ with probability $\cfrac{8}{9}$ or at point $II$ with probability $\cfrac{1}{9}$. We have $21$ shells each of which can be fired at point $I$ or $II$. Each shell may hit the target independently of the other shell with probability $\cfrac{1}{2}$. How many shells must be fired at point $I$ to hit the target with maximum probability?

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

  1. $P(A)$ is maximum where $x=11$.

  2. $P(A)$ is maximum where $x=12$.

  3. $P(A)$ is maximum where $x=14$.

  4. $P(A)$ is maximum where $x=15$.

Show answer
Correct Option: B
Explanation:

Let $A$ denote the event that the target is hit when $x$ shells are fired at point $I$.
Let ${ E } _{ 1 }$ and ${ E } _{ 2 }$ denote the events hitting $I$ and $II$, respectively
$\displaystyle \therefore P\left( { E } _{ 1 } \right) =\frac { 8 }{ 9 } ,P\left( { E } _{ 2 } \right) =\frac { 1 }{ 9 } $
Now $\displaystyle P\left( \frac { A }{ { E } _{ 1 } }  \right) =1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ x }$ and $\displaystyle P\left( \frac { A }{ { E } _{ 2 } }  \right) =1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x }$
Hence $\displaystyle P\left( A \right) =\frac { 8 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ x } \right] +\frac { 1 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x } \right] $
$\displaystyle \therefore \frac { dP\left( A \right)  }{ dx } =\frac { 8 }{ 9 } \left[ { \left( \frac { 1 }{ 2 }  \right)  }^{ x }\log { 2 }  \right] +\frac { 1 }{ 9 } \left[ -{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x }\log { 2 }  \right] $
For maximum probability $\displaystyle \frac { dP\left( A \right)  }{ dx } =0$
$\therefore x=12$   $\left[ \because { 2 }^{ 3-x }={ 2 }^{ x-21 }\Rightarrow 3-x=x-21 \right] $
Since $\displaystyle \frac { { d }^{ 2 }P\left( A \right)  }{ dx^{ 2 } } <0$ for $x=12$
$\therefore P\left( A \right) $ is maximum for $x=12$

In an entrance test, there are multiple choice questions. There are four possible options of which one is correct. The probability that a student knows the answer to a question is $90$%. If he gets the correct answer to a question, then the probability that he was guessing is

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

  1. $\displaystyle \frac { 1 }{ 37 } $

  2. $\displaystyle \frac { 36 }{ 37 } $

  3. $\displaystyle \frac { 1 }{ 4 } $

  4. $\displaystyle \frac { 1 }{ 49 } $

Show answer
Correct Option: A
Explanation:

We define the following events
${ A } _{ 1 }:$ He knows the answer
${ A } _{ 2 }:$ He does not know the answer
$E:$ He gets the correct answer
Thus $\displaystyle P\left( { A } _{ 1 } \right) =\frac { 9 }{ 10 } ,P\left( { A } _{ 2 } \right) =1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } ,P\left( \frac { E }{ { A } _{ 1 } }  \right) =1,P\left( \frac { E }{ { A } _{ 2 } }  \right) =\frac { 1 }{ 4 } $
$\therefore$ required probability $\displaystyle =P\left( \frac { { A } _{ 2 } }{ E }  \right) =\frac { P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right)  }{ P\left( { A } _{ 1 } \right) P\left( \frac { E }{ { A } _{ 1 } }  \right) +P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right)  } $
$\displaystyle =\frac { \dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  }{ \dfrac { 9 }{ 10 } .1+\dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  } =\frac { 1 }{ 36+1 } =\frac { 1 }{ 37 } $

$A$ is one of $6$ horses entered for a race, and is to be ridden by one of two jockeys $B$ and $C$. It is $2$ to $1$ that $B$ rides $A$, in which case all the horses are equally likely to win; if $C$ rides $A$, his chance is trebled; what are the odds against his winning?

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

  1. $13:5$

  2. $13:18$

  3. $18:13$

  4. $5:13$

Show answer
Correct Option: A
Explanation:
Let $E _{1}$ be the event that $B$ rides $A$, $E _{2}$, the event that $C$ rides $A$ and $E$ the event that $A$ wins. 
Then according to the question, $\displaystyle P(E _{1})=\dfrac{2}{3}, P(E _{2})=1-\dfrac{2}{3}=\dfrac{1}{3} P(E/E _{1})=\dfrac{1}{6}$ 
(since all the $6$ horses are equally likely to win when $B$ rides $A$)
$P(E/E _{2})=3\times \dfrac{1}{6}=\dfrac{1}{2}$ 
(since $A$'s chance of winning is trebled when $C$ rides $A$) 
$\displaystyle \therefore P(E)=P(E _{1})P(E/E _{1})+P(E _{2})P(E/E _{2})=\dfrac{2}{3}\cdot \dfrac{1}{6}+\dfrac{1}{3}\cdot \dfrac{1}{2}=\dfrac{58}{18}$ 
so that odds against $A$'s win are as $ (18-5):5$, that is $13:5$.

An employer sends a letter to his employee but he does not receive the reply (It is certain that employee would have replied if he did receive the letter). It is known that one out of $n$ letters does not reach its destination. Find the probability that employee does not receive the letter.

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

  1. $\displaystyle \frac{1}{n-1}.$

  2. $\displaystyle \frac{n}{2n-1}.$

  3. $\displaystyle \frac{n-1}{2n-1}.$

  4. $\displaystyle \frac{n-2}{n-1}.$

Show answer
Correct Option: C
Explanation:

Let $E$ be the event that employee received the letter and $A$ that employer received the reply, then
$\displaystyle P\left ( E \right )= \frac{n-1}{n}$ and $\displaystyle P\left ( \bar{E} \right )= \frac{1}{n}$

$\displaystyle P\left ( A/E \right )= \frac{n-1}{n}$ and $\displaystyle P\left ( A/\bar{E} \right )= 0$
Now $\displaystyle P\left ( A \right )= P\left ( E\cap A \right )+P\left ( \bar{E}\cap A \right )$
$\displaystyle = P\left ( E \right ).P\left ( A/E \right )+P\left ( \bar{E} \right ).P\left ( A/\bar{E} \right )$
$\displaystyle = \left ( \frac{n-1}{n} \right )\left ( \frac{n-1}{n} \right )+\frac{1}{n}.0$
$\displaystyle P\left ( A \right )= \left ( \frac{n-1}{n} \right )^{2}$
$\displaystyle
P\left ( \bar{A} \right )= 1-\left ( \frac{n-1}{n} \right )^{2}=
\frac{n^{2}-n^{2}-1+2n}{n^{2}}= \frac{2n-1}{n^{2}}$
Now the required probability
$\displaystyle P\left ( E/\bar{A} \right )= \frac{P\left ( E\cap \bar{A} \right  )}{P\left ( \bar{A} \right )}= \frac{P\left ( E \right )-P\left ( E\cap A

\right )}{P\left ( \bar{A} \right )}$

$\displaystyle = \frac{P\left ( E \right )-P\left ( E \right ).P\left ( A/E \right )}{P\left ( \bar{A} \right )}$
Putting the values, we get
$\displaystyle = \dfrac{\dfrac{n-1}{n}-\dfrac{n-1}{n}.\dfrac{n-1}{n}}{\dfrac{2n-1}{n^{2}}}$
$\displaystyle \therefore P\left ( E/\bar{A} \right )= \frac{n-1}{2n-1}.$

There are two groups of subjects one of which consists of 5 science subjects and 3 engineering subjects and the other consists of 3 science and 5 engineering subjects. An unbaised die is cast. If number 3 or number 5 turns up, a subject is selected at random from the first group, other wise the subject is selected at random from the second group. Find the probability that an engineering subject is selected ultimately.

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

  1. $\displaystyle \frac{13}{24}$

  2. $\displaystyle \frac{1}{3}$

  3. $\displaystyle \frac{2}{3}$

  4. $\displaystyle \frac{11}{24}$

Show answer
Correct Option: A
Explanation:

Let  $\displaystyle E _{1}$ be the event that a subject is selected from first group.
$\displaystyle E _{2}$ the event that a subject is selected from the second group.
$E$ be the event that an engineering subject is selected.
Now the probability that die shows $3$ or $5$  is
$\displaystyle P(E _1)=\frac{2}{6}=\frac{1}{3}$

$\displaystyle P\left ( E _{2} \right )=\frac{1}{3}=\frac{2}{3}.$
Now probability of choosing an engineering subject from first group is 
$\displaystyle P\left ( E|E _{1} \right )=$  $\displaystyle \frac{^{3}C _{1}}{^{8}C _{1}}=\frac{3}{8}$
Similarly, $\displaystyle P\left( E|E _{2} \right )=\frac{^{5}C _{1}}{^{8}C _{1}}=\frac{5}{8}$ 
Hence $\displaystyle P\left ( E \right )=P\left ( E _{1} \right )P\left ( E|E _{1} \right )+P\left ( E _{2}\right )P\left ( E|E _{2} \right )$
$\displaystyle =\frac{1}{3}.\frac{3}{8}+\frac{2}{3}.\frac{5}{8}$

$=\dfrac{13}{24}$