Tag: baye's theorem

Questions Related to baye's theorem

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

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

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 $x=12$