Tag: bayes theorem

Quizzes (0) Questions (62)

Questions Related to bayes theorem

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

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.

  1. $\displaystyle \frac{1}{6}$
  2. $\displaystyle \frac{15}{19}$
  3. $\displaystyle \frac{4}{19}$
  4. $\displaystyle \frac{10}{27}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac{10}{27}.$

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

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

  1. $\dfrac {1}{7}$
  2. $\dfrac {5}{13}$
  3. $\dfrac {5}{14}$
  4. $\dfrac {5}{7}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

  1. $\displaystyle \frac{24}{55}$
  2. $\displaystyle \frac{27}{55}$
  3. $\displaystyle \frac{9}{23}$
  4. $\displaystyle \frac{10}{23}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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}$

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

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?

  1. $\displaystyle \frac{4}{23}$
  2. $\displaystyle \frac{6}{23}$
  3. $\displaystyle \frac{8}{23}$
  4. $\displaystyle \frac{17}{23}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

$P(E)=\dfrac{6}{23}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

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:

  1. $\dfrac{3}{8}$
  2. $\dfrac{1}{5}$
  3. $\dfrac{3}{4}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
$P(S _1/E)=\dfrac {P(S _1).P(E/S _1)}{P(S _1)P(E/S _1)+P(S _2).P(E/S _2)}\\\implies = \dfrac {\dfrac 16\times \dfrac 34}{\dfrac 16\times \dfrac 34+\dfrac 56\times \dfrac 14}=\dfrac 18\times \dfrac {24}{8}=\dfrac {3}{8}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

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

  1. $\cfrac { 1 }{ 5 } $
  2. $\cfrac { 8 }{ 11 } $
  3. $\cfrac { 4 }{ 7 } $
  4. $\cfrac { 3 }{ 4 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$
$P(A|B)=\dfrac{0.2}{0.35}$
$P(A|B)=\dfrac{4}{7}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

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:

  1. $n+1C _{2}$
  2. $(2)(n\ !)^{2}$
  3. $\dfrac{(n\ !)}{2}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 =$2{ (n!) }^{ 2 }$

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

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?

  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 $x=15$.
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $x=12$

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

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

  1. $\displaystyle \frac { 1 }{ 37 } $
  2. $\displaystyle \frac { 36 }{ 37 } $
  3. $\displaystyle \frac { 1 }{ 4 } $
  4. $\displaystyle \frac { 1 }{ 49 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $

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

$A$ 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?

  1. $13:5$
  2. $13:18$
  3. $18:13$
  4. $5:13$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $13:5$.