Tag: baye's theorem

Questions Related to baye's theorem

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 } $

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
$\displaystyle P\left( \frac { A }{ B }  \right) =\frac { P\left( A \right) P\left( \frac { B }{ A }  \right)  }{ P\left( A \right) P\left( \frac { B }{ A }  \right) +P\left( A' \right) P\left( \frac { B }{ A' }  \right)  } =\frac { 0.32 }{ 0.32+0.15 } =\frac { 32 }{ 47 } =68$%