Tag: probability and probability distribution

Questions Related to 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.

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 $x=12$

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 $13:5$.