Tag: business maths

Questions Related to business maths

A bag contains 12 balls out of which x are white.If one ball is drawn at random, what is the probability it will be a white ball?

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

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

  2. $\displaystyle \frac{x}{12}$

  3. $ \displaystyle \frac{x}{10}$

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

Show answer
Correct Option: B
Explanation:

Total number of balls = 12
Number of white balls = x
P (white ball) = $= \displaystyle \frac{x}{12}$

A pack of playing cards was found to contain only $51$ cards. If the first $13$ cards which are examined are all red, then the probability thatthe missing card is black, is

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

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

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

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

  4. $\displaystyle \frac{^{25} C _{13}}{^{51} C _{13}}$

Show answer
Correct Option: B
Explanation:

$Total\quad number\quad of\quad cards=52\ Number\quad of\quad lost\quad cards=1\ 13\quad cards\quad are\quad surely\quad red\quad therfore,\quad from\quad the\quad remaining\quad 39\quad cards\quad 26\quad are\quad black\quad and\quad 13\quad are\quad red\ So\quad probability\quad of\quad lost\quad card\quad being\quad black\quad =\frac { \left( 26\ 1 \right)  }{ \left( 39\ 1 \right)  } =\frac { 26 }{ 39 } =\frac { 2 }{ 3 } $

Bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bag is found to be red. Find the probability that it was drawn from bag $B$.

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

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

  2. $\dfrac{25}{52}$

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

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

Show answer
Correct Option: B
Explanation:

Let $X$ be the probability of choosing bag $A$,$Y$ be the probability of choosing bag $B$


Let $E$ be the probability of ball drawn is red


Then $P\left( X \right) = \dfrac{1}{2}$

$P\left( Y \right) = \dfrac{1}{2}$

$P\left( {E/X} \right) = \dfrac{3}{5}$

$P\left( {E/Y} \right) = \dfrac{5}{9}$

Apply the Bayes theorem:

$P\left( {Y/E} \right) = \dfrac{{P\left( Y \right) \times P\left( {E/Y} \right)}}{{P\left( X \right) \times P\left( {E/X} \right) + P\left( Y \right) \times P\left( {E/Y} \right)}}$

                 $ = \dfrac{{\dfrac{1}{2} \times \dfrac{5}{9}}}{{\dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{5}{9}}}$

                 $ = \dfrac{{50}}{{104}}$

                 $ = \dfrac{{25}}{{52}}$

Hence, the probability that the red ball is drawn from bag $B$ is  $ = \dfrac{{25}}{{52}}$

An urn contains $10$ balls coloured either black or red When selecting two balls from the urn at random, the probability that a ball of each color is selected is $8/15$. Assuming that the urn contains more black balls then red balls, the probability that at least one black ball is selected, when selecting two balls, is

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

  1. $\dfrac {18}{45}$

  2. $\dfrac {30}{45}$

  3. $\dfrac {39}{45}$

  4. $\dfrac {41}{45}$

Show answer
Correct Option: C
Explanation:
Number of black balls $=x$

Number of red balls $=y$

$i)=x+y=10$ (Total)

$ii)\ P$ (selecting exactly $1$ black & one red)

$=^xC _1\times ^yC _1 /^{10}C _2=8/15$

by equation : $x^2-10x+24=0\ ;\ x=4$ or $6$

since $ x > y$ it is $6$

$iii)\ P$ (selecting at least one black)

$=^xC _1\times ^yC _1 +^xC _2 /^{10}C _2$

Above equation reduced to $\Rightarrow \ 2xy+x(x-1)/90$

putting $x=6$, results is $39/45$

There are six letters $L _1, L _2, L _3, L _4, L _5, L _6$ are their corresponding six envelopes $E _1, E _2, E _3, E _4, E _5, E _6$. Letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes, then number of arrangement equals?

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

  1. $6$

  2. $9$

  3. $44$

  4. $4$

Show answer
Correct Option: D
Explanation:

There are three odd envelops and three even envelops 

Favourable ways so that  letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes are
For odd 1,3,5 -,,_
5 1 3,3 5 1

For even 2,4,6- _,_,_
6 2 4,4 6 2

There are four ways

Two unbiased dice are thrown. The probability that the sum of the numbers appearing on the top face of two dice is greater than $7$ if $4$ appear on the top face of the first dice is...

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

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

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

  3. $\dfrac {1}{12}$

  4. $\dfrac {1}{6}$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l}\left. \begin{array}{l}{\rm{combinations}} = \left( {4,1} \right)\left( {4,2} \right)\\left( {4,3} \right)\left( {4,4} \right)\end{array} \right} \le 7\\left. {\left( {4,5} \right)\left( {4,6} \right)} \right} > 7\{\rm{P}}\left( { > 7} \right) = \frac{1}{3}\end{array}$

In a class 5% of boys and 10% of girls have an I.Q of more than 150.In this class 60% of students are boys. If a student is selected at random and found to have an I.Q. of more than 150. Find the probability that the student is a boy.

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

  1. $\dfrac{3}{7}$

  2. $\dfrac{23}{7}$

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

  4. None of these

Show answer
Correct Option: A
Explanation:
Let us consider the problem
$E _1$ : Event that boys are selected
$E _2$ : Event that girls are selected
$A$ : event that have IQ $150$
Implies that,
\begin{array}{l} P\left( { A/{ E_{ 1 } } } \right) =\dfrac { 5 }{ { 100 } }  \\ P\left( { A/{ E_{ 2 } } } \right) =\dfrac { { 10 } }{ { 100 } }  \\ P\left( { { E_{ 1 } } } \right) =\dfrac { { 60 } }{ { 100 } } ,P\left( { { E_{ 2 } } } \right) =\dfrac { { 40 } }{ { 100 } }  \\ P\left( { A|{ E_{ 1 } } } \right) =\dfrac { { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right)  } }{ { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right) +P\left( { { E_{ 2 } } } \right) P\left( { A|{ E_{ 2 } } } \right)  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } }  } }{ { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } +\dfrac { { 40 } }{ { 100 } } \times \dfrac { { 10 } }{ { 100 } }  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 60\times 5 } }{ { 60\times 5+40\times 10 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 300 } }{ { 300+400 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { 3 }{ 7 }  \end{array}

Hence, the probability is $\dfrac {3}{7}$

A bag contains $6$ red, $4$ white and $8$ blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.

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

  1. $\dfrac {2}{17}$

  2. $\dfrac {3}{17}$

  3. $\dfrac {5}{17}$

  4. $\dfrac {4}{17}$

Show answer
Correct Option: D
Explanation:
$E\rightarrow$ Event of getting one is red, one is white and one $6$ red $+4$ white $+8$ blue balls $=18$ balls

Total outcomes $=(\ ^{18}C _{3})$

$\underbrace { \bigodot  } _{ R } \underbrace { \bigodot  } _{ W } \underbrace { \bigodot  } _{ B } \rightarrow$ no. of fobourable element

$=\ ^{6}C _{1}\times \ ^{4}C _{1}\times \ ^{8}C _{1}$

$=6\times 4\times 8$

$\therefore P(E)=\dfrac{6\times 4\times 8}{\ ^{18}C _{3}}=\dfrac{6\times 4\times 8\times 3\times 2\times 1}{18\times 17\times 16}$

$=\dfrac{4}{17}$

There are two balls in an urn whose colors are not known ( ball can be either white or black). A white ball is put into the urn. A ball is then drawn from the urn. The probability that it is white is 

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

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

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

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

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

Show answer
Correct Option: C
Explanation:

Let $\displaystyle { E } _{ i }\left( 0\le i\le 2 \right) $ denotes the event that urn contains $i$ white and $2-i$ black balls.

Let $A$ denotes the event that a white ball is drawn from the urn.
We have $\displaystyle P\left( { E } _{ i } \right) =\frac { 1 }{ 3 } $ for $i=0,1,2$ and $\displaystyle P\left( \frac { A }{ { E } _{ i } }  \right) =\frac { 1 }{ 3 } ,P\left( \frac { A }{ { E } _{ 2 } }  \right) =\frac { 2 }{ 3 } ,P\left( \frac { A }{ { E } _{ 3 } }  \right) =1$
By the total probability rule,
$\displaystyle P\left( A \right) =P\left( { E } _{ 1 } \right) P\left( \frac { A }{ { E } _{ 1 } }  \right) +P\left( { E } _{ 2 } \right) P\left( \frac { A }{ { E } _{ 2 } }  \right) +P\left( { E } _{ 3 } \right) P\left( \frac { A }{ { E } _{ 3 } }  \right) $
$\displaystyle =\frac { 1 }{ 3 } \left[ \frac { 1 }{ 3 } +\frac { 2 }{ 3 } +1 \right] =\frac { 2 }{ 3 } $

Cards are dealt one by one from a well shuffled pack until an ace appears. the probability that exactly n cards are dealt befor  the first ace appears is

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

  1. $ \dfrac { 4\left( 51-n \right) \left( 50-n \right) \left( 49-n \right) }{ 52.51.50.49 } $

  2. $ \dfrac { 4\left( 52-n \right) \left( 51-n \right) \left( 49-n \right) }{ 52.51.50.49 } $

  3. $ \dfrac { 4\left( 52-n \right) \left( 51-n \right) \left( 49-n \right) }{ 51.50.49.48 } $

  4. $ \dfrac { 4\left( 51-n \right) \left( 50-n \right) \left( 49-n \right) }{ 51.50.49.48 } $

Show answer
Correct Option: A
Explanation:

A: number of ace is drawn in the first n draw
B: an ace appear in the ${ \left( n+1 \right)  }^{ th }$ draw
Hence the probability that exactly n cards are dealt before the first ace appear is equal to $P\left( A\cap B \right) $
$\displaystyle P\left( A \right) =\frac { ^{ 48 }{ { C } _{ n } } }{ ^{ 52 }{ { C } _{ n } } } ,P\left( \frac { B }{ A }  \right) =\frac { 4 }{ 52-n } $
$\displaystyle \therefore P\left( A\cap B \right) =P\left( A \right) .P\left( \frac { B }{ A }  \right) =\frac { ^{ 48 }{ { C } _{ n } } }{ ^{ 52 }{ { C } _{ n } } } .\frac { 4 }{ 52-n } $
$\displaystyle =\frac { 48! }{ \left( 48-n \right) !n! } \times \frac { \left( 52-n \right) !n! }{ 52! } \times \frac { 4 }{ 52-n } $
$\displaystyle =\frac { 4\left( 51-n \right) \left( 50-n \right) \left( 49-n \right)  }{ 52\times 51\times 50\times 49 } $