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Questions Related to business maths

There are $3$ coins in a box. One is a two-headed coin; another is a fair coin; and third is biased coin that comes up heads $75\%$ of time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that its was the two-headed coin ?

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

  1. $\dfrac{2}{9}$

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

  3. $\dfrac{4}{9}$

  4. $\dfrac{5}{9}$

Show answer
Correct Option: C
Explanation:
Formula using Baye's theorem,

$P(E _1|A)=\dfrac{P(E _1)P(A|E _1)}{P(E _1)P(A|E _1)+P(E _2)P(A|E _2)+P(E _3)P(A|E _3)}$

Let E1: event that the coin is 2 headed, 

E2 be the event that its biased with heads 75% of the time and 

E3 be the fair coin.

All these three events are mutually exclusive and exchaustive, and are equally likely.

$\therefore P(E _1)=P(E _2)=P(E _3)=\dfrac{1}{3}$

P( coin shows head given that its 2 headed coin)$ =P(E|E _1)=1$

P(coin shows head given that its 75% biased for heads) $=P(E|E _2)=\dfrac{3}{4}$

P(coin shows head given that its a fair coin) $=P(E|E3)=\dfrac{1}{2}$

$\therefore P(E _1|E)=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{3}{4} +\dfrac{1}{3}\cdot \dfrac{1}{2}}$

$P(E _1|E)=\dfrac{4}{9}$

If in Q. 104, we are told that a white ball has been drawn, find the probability that it was drawn from the first urn.

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

  1. $\displaystyle \frac{5}{9}.$

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

  3. $\displaystyle \frac{2}{9}.$

  4. $\displaystyle \frac{7}{9}.$

Show answer
Correct Option: C
Explanation:

Here we have to find $\displaystyle P\left ( A _{1}/B \right ).$
By Baye's theorem
$\displaystyle P\left ( A _{1}/B \right )= \frac{P\left ( A _{1} \right )P\left ( B/A _{1} \right )}{P\left ( A _{1} \right )P\left ( B/A _{1} \right )+P\left ( A _{2} \right )P\left ( B/A _{2} \right )+P\left ( A _{3} \right )P\left ( B/A _{3} \right )}$
$\displaystyle = \dfrac{\dfrac{1}{3}.\dfrac{2}{5}}{\dfrac{3}{5}},$
$\displaystyle = \frac{2}{9}.$

A letter is known to have come eithe from London or Clifton; on the post only the consecutive letters ON are legible; what is the chance that it came from London?

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

  1. $\displaystyle \frac{12}{17}$

  2. $\displaystyle \frac{5}{17}$

  3. $\displaystyle \frac{5}{12}$

  4. $\displaystyle \frac{7}{12}$

Show answer
Correct Option: A
Explanation:

If letter came from Clifton there are $6$ pairs of consecutive letters i.e., $cl, li, if, ft, to$, $ON $ in which $ON$ appears only once.
$\displaystyle \therefore $ the chance that this was the legible couple on the Clifton hypothesis $\displaystyle = \frac{1}{6}$
pairs of consecutive letters in the word London are $lo, on, nd, do$, $ON $ in which $ON$ occurs twice.
$\displaystyle \therefore $ the chance that this was the legible couple on the London hypothesis$=2/5.$
$\displaystyle \therefore $ The a posteriori chances that the letter was from Clifton or London are $\displaystyle \frac{1/6}{\dfrac{1}{6}+\dfrac{2}{5}}$ and $\displaystyle \frac{2/5}{\dfrac{1}{6}+\dfrac{2}{5}}$ respectively.
Thus the reqd.chance $\displaystyle = \frac{12}{17}.$

A person is know to speak the truth 4 times out of 5. He throws a die and reports that it is a ace. The probability that it is actually a ace is

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

  1. $1/3$

  2. $2/9$

  3. $4/9$

  4. $5/9$

Show answer
Correct Option: C
Explanation:

Let $E _1$ denote the event that an ace occurs and $E _2$ the event that it does not occur. Let $A$ denote the event that the person reports that it is an ace. 

Then $P(E _1)=1/6, P(E _2)=5/6, P(A|E _1)=4/5$ an $P(A|E _2)=1/5$. 
By Bayes' theorem,
$P(E _1|A)=\dfrac {P(E _1)P(A|E _1)}{P(E _1)P(A|E _1)+P(E _2)P(A|E _2)}$
$=\dfrac {4}{9}$

A is known to tell the truth in $5$ cases out of $6$ and he states that a white ball was drawn from a bag containing $8$ black and $1$ white ball. The probability that the white ball was drawn, is

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

  1. $\displaystyle \frac { 7 }{ 13 } $

  2. $\displaystyle \frac { 5 }{ 13 } $

  3. $\displaystyle \frac { 9 }{ 13 } $

  4. None of these

Show answer
Correct Option: B
Explanation:

Let $W$ denote the event that $A$ draws a white ball and $T$ the event that $A$ speak truth.
In the usual notations, we are given that 
$\displaystyle P\left( W \right) =\frac { 1 }{ 9 } ,P\left( \frac { T }{ w }  \right) =\frac { 5 }{ 6 } $
so that $\displaystyle P\left( \overline { W }  \right) =1-\frac { 1 }{ 9 } =\frac { 8 }{ 9 } ,P\left( \frac { T }{ \overline { W }  }  \right) =1-\frac { 5 }{ 6 } =\frac { 1 }{ 6 } $.
Using Baye's theorem required probability is given by 
$\displaystyle P\left( \frac { W }{ T }  \right) =\frac { P\left( W\cap T \right)  }{ P\left( T \right)  } =\frac { P\left( W \right) P\left( \frac { T }{ w }  \right)  }{ P\left( W \right) P\left( \frac { T }{ w }  \right) +P\left( \overline { W }  \right) P\left( \frac { T }{ \overline { W }  }  \right)  } $
$\displaystyle =\frac { \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 }  }{ \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 } +\dfrac { 8 }{ 9 } \times \dfrac { 1 }{ 6 }  } =\frac { 5 }{ 13 } $

At the college entrance examination each candidate is admitted or rejected according to whether he has passed or failed the tests. Of the candidate who are really capable, $80$% pass the test and of the incapable, $25$% pass the test. Given that $40$% of the candidates are really capable, then the proportion of capable college students is about 

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

  1. $68$%

  2. $70$%

  3. $73$%

  4. $75$%

Show answer
Correct Option: A
Explanation:

Let $A$ be the event that a really able candidate passes the test 
and let $B$ be the event that any candidate passes this test.
Then we have
$\displaystyle P\left( \frac { B }{ A }  \right) =0.8,P\left( \frac { B }{ A' }  \right) =0.25,P\left( A \right) =0.4,P\left( A' \right) =1-0.4=0.6$
By Baye's formula
$\displaystyle P\left( \frac { A }{ B }  \right) =\frac { P\left( A \right) P\left( \frac { B }{ A }  \right)  }{ P\left( A \right) P\left( \frac { B }{ A }  \right) +P\left( A' \right) P\left( \frac { B }{ A' }  \right)  } =\frac { 0.32 }{ 0.32+0.15 } =\frac { 32 }{ 47 } =68$%

A box has four dice in it. Three of them are fair dice but the fourth one has the number five on all of its faces. A die is chosen at random from the box and is rolled three times and shows up the face five on all the three occasions. The chance that the die chosen was a rigged die, is

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

  1. $\displaystyle \frac {216}{217}$

  2. $\displaystyle \frac {215}{219}$

  3. $\displaystyle \frac {216}{219}$

  4. none

Show answer
Correct Option: C
Explanation:

Given, $A$ is the event which selects the rigged one and $B$ is the event which selects the fair one.
Let E is the event which shows 5 in all three times
Probability of event A for given event E, $P(A/E)=\dfrac{1}{4}(1)^3=\dfrac{1}{4}$ ($\dfrac{1}{4}$ in the equation is probability of selecting one dice among 4)
Probability of event B for given event E, $P(B/E)=\dfrac{3}{4}(\dfrac{1}{6})^3=\dfrac{1}{1152}$ (since probability of getting 5 in fair dice case=$\dfrac{1}{5}$)
By Baye's theorem,Probability of selecting the rigged case among both=$\dfrac{P(A/E)}{P(A/E)+P(B/E)}=\dfrac{(\dfrac{1}{4})}{(\dfrac{1}{4})+(\dfrac{1}{1152})}=\dfrac{216}{219}$

Suppose that of all used cars of a particular year 30% have bad brakes. You are considering buying a used car of that year. You take the car to a mechanic to have the brakes checked. The chance that the mechanic will give you the wrong report is 20%. Assuming that the car you take to the mechanic is selected at random from the population of cars of that year. The chance that the car's brakes are good, given that the mechanic says its brakes are good, is

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

  1. $\displaystyle \frac{28}{130}$

  2. $\displaystyle \frac{29}{31}$

  3. $\displaystyle \frac{37}{62}$

  4. $\displaystyle \frac{29}{62}$

Show answer
Correct Option: A
Explanation:
Given $30$% of the cars of bad brakes
$P(E _1)=70$%$=\dfrac7{10}$        $P(E _2)=30$%$=\dfrac3{10}$
$\Rightarrow P\left( \dfrac { { E } }{ { E } _{ 1 } }  \right) =0.2\times0.2=0.04$
$\Rightarrow P\left( \dfrac { { E } }{ { E } _{ 2 } }  \right) =0.8\times0.8=0.64$
$\therefore P\left( \dfrac { { E } }{ { E } _{ 1 } }  \right) =\dfrac { \dfrac { 7 }{ 10 } \times \dfrac { 2 }{ 10 } \times \dfrac { 2 }{ 10 }  }{ \dfrac { 7\times 4 }{ 1000 } +\dfrac { 3 }{ 10 } +\dfrac { 8 }{ 10 } +\dfrac { 8 }{ 10 }  } =\dfrac { 28 }{ 102+28 } =\dfrac { 28 }{ 130 } $
Hence, the answer is $\dfrac { 28 }{ 130}.$

Box $I$ contains $5$ red and $4$ blue balls, while box $II$ contains $4$ red and $2$ blue balls. A fair die is thrown. If it turns up a multiple of $3$, a ball is drawn from the box $I$ else a ball is drawn from box $II$. Find the probability of the event ball drawn is from the box $I$ if it is blue.

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

  1. $\displaystyle \frac{1}{6}$

  2. $\displaystyle \frac{15}{19}$

  3. $\displaystyle \frac{4}{19}$

  4. $\displaystyle \frac{10}{27}$

Show answer
Correct Option: D
Explanation:
Box $I$ contains $:5$ red $+{4}$ blue
Box $II$ contains $:4$ red $+{2}$ blue
A ball is taken from Box $I$ if a multiple of $3$ comes up i.e, $3$ and $6.$

Ball is taken from Box $II$ when $1,2,4$ and $5$ turns up.

$\Rightarrow$ Event of picking up from Box $I=P(A _1)=\dfrac{2}{6}=\dfrac{1}{3}.$

$\Rightarrow$ Event of picking up from Box $II=P(A _2)=\dfrac{4}{6}=\dfrac{2}{3}.$

$\Rightarrow R=$ event of drawing a blue ball

$=P(A _1)P(\dfrac R{A _1})+P(A _2)P(\dfrac{R}A _2)$

$=\dfrac{1}{3}\times\dfrac{4}{9}+\dfrac{2}{3}\times\dfrac{2}{6}$

$=\dfrac{4}{27}+\dfrac{4}{18}$

$=\dfrac{10}{27}.$
Hence, the answer is $\dfrac{10}{27}.$

There are three different Urns, Urn-I, Urn-II and Urn-III containing 1 Blue, 2 Green, 2 Blue, 1 Green, 3 Blue, 3 Green balls respectively. If two Urns are randomly selected and a ball is drawn from each Urn and if the drawn balls are of different colours then the probability that chosen Urn was Urn-I and Urn-II is

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

  1. $\dfrac {1}{7}$

  2. $\dfrac {5}{13}$

  3. $\dfrac {5}{14}$

  4. $\dfrac {5}{7}$

Show answer
Correct Option: C
Explanation:

Required probability$\displaystyle =\dfrac {\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {1}{3}+\dfrac {2}{3}.\dfrac {2}{3}\right )}{\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {2}{3}.\dfrac {2}{3}\right )+\dfrac {1}{3}\left (\dfrac {2}{3}.\dfrac {3}{6}+\dfrac {1}{3}.\dfrac {3}{6}\right )+\dfrac {1}{3}\left (\dfrac {3}{6}.\dfrac {2}{3}+\dfrac {3}{6}.\dfrac {1}{3}\right )}$

$\displaystyle =\dfrac {\dfrac {5}{9}}{\dfrac {5}{9}+\dfrac {9}{18}+\dfrac {9}{18}}\\ =\dfrac {5}{14}$