Probability Test 21

Possible cases are: 
(i) $$3B, 7W$$; (ii) $$4B, 6W$$; (iii) $$5B, 5W$$; (iv) $$6B, 4W$$;
(v) $$7B, 3W$$; (vi) $$8B, 2W$$; (vii) $$9B, 1W$$; (viii) $$10B, 0W$$
Probability of each case $$=\dfrac { 1 }{ 8 } $$
Probability of getting $$3B$$ in the above cases:
(i) $$\dfrac { ^{ 3 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (ii) $$\dfrac { ^{ 4 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (iii) $$\dfrac { ^{ 5 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (iv) $$\dfrac { ^{ 6 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$

(v) $$\dfrac { ^{ 7 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (vi) $$\dfrac { ^{ 8 }C_{ 3 } }{ ^{ 10 }C_{ 3 } }$$; (vii) $$\dfrac { ^{ 9 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (viii) $$\dfrac { ^{ 10 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$

Therefore, required probability is
$$ \displaystyle \dfrac { \dfrac { ^{ 9 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } \times \dfrac { 1 }{ 8 }  }{ \left( \dfrac { ^{ 3 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 4 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 5 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 6 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 7 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 8 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 9 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 10 }C_{ 3 } }{ ^{ 10 }C_{ 3 } }  \right) \dfrac { 1 }{ 8 }  } =\dfrac { 84 }{ 330 } =\dfrac { 14 }{ 55 } $$

Hence, A is correct.

Probability that man speaks the truth is $$\dfrac{3}{4}$$.

The probability that man lies is $$\dfrac{1}{4}$$.

Probability of getting a 6 $$=\dfrac{1}{6}$$

Probability of not getting a six $$=\dfrac{5}{6}$$

Applying Baye's theorem, we get the required probability as 
$$\dfrac{\dfrac{1}{6}.\dfrac{3}{4}}{\dfrac{1}{6}.\dfrac{3}{4}+\dfrac{5}{6}.\dfrac{1}{4}}$$

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

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 $$\displaystyle =P\left( \frac { { A }_{ 2 } }{ E }  \right) =\frac { P\left( { A }_{ 2 } \right) P\left( \frac { E }{ { A }_{ 2 } }  \right)  }{ P\left( { A }_{ 1 } \right) P\left( \frac { E }{ { A }_{ 1 } }  \right) +P\left( { A }_{ 2 } \right) P\left( \frac { E }{ { A }_{ 2 } }  \right)  } $$
$$\displaystyle =\frac { \dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  }{ \dfrac { 9 }{ 10 } .1+\dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  } =\frac { 1 }{ 36+1 } =\frac { 1 }{ 37 } $$

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 $$x=12$$

The event of selecting a red ball is denoted by 'R'.
The event of selecting the bag A is denoted by 'A'.
The event of selecting the bag B is denoted by 'B'.

$$P(A)=0.25$$
$$P(B)=0.35$$
$$P(C)=0.40$$
Hence
Applying baye's theorem, we get to find the possibility that the  bolt was manufactured by A is
$$=\dfrac{0.25\times 0.05}{(0.25(0.05))+(0.35(0.04))+(0.4(0.02) )}$$

$$=\dfrac{0.0125}{0.0345}$$

$$=\dfrac{25}{69}$$
Similarly 
Applying baye's theorem, we get to find the possibility that the  bolt was manufactured by B is 
$$=\dfrac{0.35\times 0.04}{(0.25(0.05))+(0.35(0.04))+(0.4(0.02) )}$$
$$=\dfrac{0.014}{0.0345}$$

$$=\dfrac{28}{69}$$
Applying baye's theorem, we get to find the possibility that the  bolt was manufactured by C is 
$$=\dfrac{0.4\times 0.02}{(0.25(0.05))+(0.35(0.04))+(0.4(0.02) )}$$
$$=\dfrac{0.008}{0.0345}$$

$$=\dfrac{16}{69}$$

Let $${ E }_{ \lambda  }$$ be the event that exactly $$\lambda $$ out of n pass the examination and 
$$A$$ is the event that a student pass the examination
$$P\left( { E }_{ \lambda  } \right) \propto { \lambda  }^{ 2 }\Rightarrow P\left( { E }_{ \lambda  } \right) ={ k\lambda  }^{ 2 }$$ (k os proportionality constant)
$$\therefore { E }_{ 0 },{ E }_{ 1 },{ E }_{ 2 },...{ E }_{ n }$$ are mutually exclusive and exhaustive events.
$$\Rightarrow P\left( { E }_{ 0 } \right) +P\left( { E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) +...P\left( { E }_{ n } \right) =1\\ \Rightarrow 0+k{ \left( 1 \right)  }^{ 2 }+k{ \left( 2 \right)  }^{ 2 }+...+k{ \left( n \right)  }^{ 2 }=1\Rightarrow k\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...{ n }^{ 2 } \right) =1$$
$$\displaystyle \Rightarrow \frac { kn\left( n+1 \right) \left( 2n+1 \right)  }{ 6 } =1\Rightarrow k=\frac { 6 }{ n\left( n+1 \right) \left( 2n+1 \right)  } $$
$$\displaystyle \\ P\left( A \right) =\sum _{ r=1 }^{ n }{ P\left( { E }_{ \lambda  } \right)  } P\left( \frac { A }{ { E }_{ \lambda  } }  \right) =\sum _{ r=1 }^{ n }{ k{ \lambda  }^{ 2 } } \times \frac { \lambda  }{ n } =\frac { k }{ n } \sum _{ r=1 }^{ n }{ { \lambda  }^{ 3 } } $$
$$\displaystyle =\frac { k }{ n } { \left[ \frac { n\left( n+1 \right)  }{ 2 }  \right]  }^{ 2 }=\frac { 6 }{ { n }^{ 2 }\left( n+1 \right) \left( 2n+1 \right)  } .\frac { { n }^{ 2 }{ \left( n+1 \right)  }^{ 2 } }{ 4 } =\frac { 3\left( n+1 \right)  }{ 2\left( 2n+1 \right)  } $$
$$\displaystyle P\left( \frac { { E }_{ 1 } }{ A }  \right) =\frac { P\left( { E }_{ 1 } \right) P\left( \frac { A }{ { E }_{ 1 } }  \right)  }{ P\left( A \right)  } =\frac { \frac { 6 }{ n\left( n+1 \right) \left( 2n+1 \right)  } \times \frac { 1 }{ n }  }{ \frac { 3\left( n+1 \right)  }{ 2\left( 2n+1 \right)  }  } =\frac { 4 }{ { \left[ n\left( n+1 \right)  \right]  }^{ 2 } } $$

Let $${ E }_{ 1 },{ E }_{ 2 },{ E }_{ 3 }$$ and $$A$$ be the events defined as:
$${ E }_{ 1 }\rightarrow $$ The examine guesses the answer.
$${ E }_{ 2 }\rightarrow $$ The examine copies the answer
$${ E }_{ 3 }\rightarrow $$ The examine knows the answer.
$$A\rightarrow $$ The examine answer correctly
We have, $$\displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ 3 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ 6 } $$
Since, $${ E }_{ 1, }{ E }_{ 2 },{ E }_{ 3 }$$ are mutually exclusive and exhaustive events.
$$\displaystyle \therefore P\left( { E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) +P\left( E_{ 3 } \right) =1\Rightarrow P\left( { E }_{ 3 } \right) =1-\frac { 1 }{ 3 } -\frac { 1 }{ 6 } =\frac { 1 }{ 2 } $$
If $${ E }_{ 1 }$$ has already occurred, then the examine gusses. Since, there are four choices out of which only one is correct, therefore the probability that he answer correctly given that he has made a guess is $$\displaystyle \frac { 1 }{ 4 } $$

i.e $$\displaystyle P\left( \frac { A }{ { E }_{ 1 } } \right) =\frac { 1 }{ 4 } $$

It is given that $$\displaystyle P\left( \frac { A }{ { E }_{ 2 } } \right) =\frac { 1 }{ 8 } $$

and $$\displaystyle P\left( \frac { A }{ { E }_{ 3 } } \right) =$$ Probability that he answer correctly given that he knows the answer =$$1$$

By Baye's theorem, we have

$$\displaystyle P\left( \frac { { E }_{ 3 } }{ A } \right) =\frac { P\left( { E }_{ 3 } \right) P\left( \frac { A }{ { E }_{ 3 } } \right) }{ \left[ P\left( { E }_{ 1 } \right) .P\left( \frac { A }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) .P\left( \frac { A }{ { E }_{ 2 } } \right) +P\left( { E }_{ 3 } \right) .P\left( \frac { A }{ { E }_{ 3 } } \right) \right] } $$

$$\displaystyle \therefore P\left( \frac { { E }_{ 3 } }{ A } \right) =\frac { \frac { 1 }{ 2 } \times 1 }{ \left( \frac { 1 }{ 3 } \times \frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 6 } \times \frac { 1 }{ 8 } \right) +\left( \frac { 1 }{ 2 } \times 1 \right) } =\frac { 24 }{ 29 } $$

Let $$E_{1}$$ be the event that die $$A$$ is used and $$E_{2}$$ be the event that die $$B$$ is used