Tag: baye's theorem

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

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.

  1. $\dfrac{3}{7}$
  2. $\dfrac{23}{7}$
  3. $\dfrac{3}{5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac {3}{7}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

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.

  1. $\dfrac {2}{17}$
  2. $\dfrac {3}{17}$
  3. $\dfrac {5}{17}$
  4. $\dfrac {4}{17}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

$=6\times 4\times 8$

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

$=\dfrac{4}{17}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

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 

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

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

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

  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 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $

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

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 ?

  1. $\dfrac{2}{9}$
  2. $\dfrac{1}{3}$
  3. $\dfrac{4}{9}$
  4. $\dfrac{5}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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) $=P(E|E3)=\dfrac{1}{2}$

$\therefore P(E _1|E)=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{3}{4} +\dfrac{1}{3}\cdot \dfrac{1}{2}}$

$P(E _1|E)=\dfrac{4}{9}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

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.

  1. $\displaystyle \frac{5}{9}.$
  2. $\displaystyle \frac{2}{3}.$
  3. $\displaystyle \frac{2}{9}.$
  4. $\displaystyle \frac{7}{9}.$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here we have to find $\displaystyle P\left ( A _{1}/B \right ).$
By Baye's theorem
$\displaystyle P\left ( A _{1}/B \right )= \frac{P\left ( A _{1} \right )P\left ( B/A _{1} \right )}{P\left ( A _{1} \right )P\left ( B/A _{1} \right )+P\left ( A _{2} \right )P\left ( B/A _{2} \right )+P\left ( A _{3} \right )P\left ( B/A _{3} \right )}$
$\displaystyle = \dfrac{\dfrac{1}{3}.\dfrac{2}{5}}{\dfrac{3}{5}},$
$\displaystyle = \frac{2}{9}.$

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

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?

  1. $\displaystyle \frac{12}{17}$
  2. $\displaystyle \frac{5}{17}$
  3. $\displaystyle \frac{5}{12}$
  4. $\displaystyle \frac{7}{12}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\displaystyle = \frac{12}{17}.$

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

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

  1. $1/3$
  2. $2/9$
  3. $4/9$
  4. $5/9$
Reveal answer Fill a bubble to check yourself
C Correct answer
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,
$P(E _1|A)=\dfrac {P(E _1)P(A|E _1)}{P(E _1)P(A|E _1)+P(E _2)P(A|E _2)}$
$=\dfrac {4}{9}$

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

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

  1. $\displaystyle \frac { 7 }{ 13 } $
  2. $\displaystyle \frac { 5 }{ 13 } $
  3. $\displaystyle \frac { 9 }{ 13 } $

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

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 

  1. $68$%
  2. $70$%
  3. $73$%
  4. $75$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$\displaystyle P\left( \frac { A }{ B }  \right) =\frac { P\left( A \right) P\left( \frac { B }{ A }  \right)  }{ P\left( A \right) P\left( \frac { B }{ A }  \right) +P\left( A' \right) P\left( \frac { B }{ A' }  \right)  } =\frac { 0.32 }{ 0.32+0.15 } =\frac { 32 }{ 47 } =68$%