Probability Test - 48

Let $$E=$$ event when each American man is seated adjacent to his wife.
$$A = $$event when Indian man is seated adjacent to his wife.
So, there are 5 couples out of which 4 are American and 1 is Indian.

Now, $$n(A\cap E)=4!(2!)^{5}$$. (Each couple is taken together as one entity, so we have total $$5$$ things to be arranged in a circle, which can be done in $$(5-1)!=4!$$ number of ways. Also, each couple as one entity can be seated in $$2$$ ways.)

And $$n(E)=5!(2!)^{4}$$ (Each American couple is taken as one entity, and an Indian man and woman separately. Hence, the arrangement of $$6$$ things in a circle$$=5!$$. Also each American couple as one entity can be seated in $$2$$ ways. )

So, $$P(A|E)=\displaystyle \frac{n(A\cap E)}{n(E)}$$
$$\Rightarrow P(A|E)=\displaystyle \frac {4!(2!)^{5}}{5!(2!)^{4}}$$
$$\Rightarrow  P(A|E)=\displaystyle \frac{2}{5}$$

Let $${ E }_{ 1 }:$$ event that one of the first $$n$$ urn is chosen 
$${ E }_{ 2 }:$$ event that $$(n+1)th$$ urn is chosen
$$A:$$ The vent that the two balls drawn from the urn without replacement are black.
Then we have $$\displaystyle P\left( { E }_{ 1 } \right) =\frac { n }{ n+1 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ n+1 } $$
$$\displaystyle P\left( \frac { A }{ { E }_{ 1 } }  \right) =\frac { _{  }^{ 6 }{ { C }_{ 2 } } }{ _{  }^{ 10 }{ { C }_{ 2 } } } =\frac { 1 }{ 3 } $$ and $$\displaystyle P\left( \frac { A }{ { E }_{ 2 } }  \right) =\frac { _{  }^{ 5 }{ { C }_{ 2 } } }{ _{  }^{ 10 }{ { C }_{ 2 } } } =\frac { 2 }{ 9 } $$

Let $$A$$ be the event that the maximum number on the two chosen tickets is not more than $$10$$, and B be the event that the maximum number on them is $$5.$$
$$\displaystyle \therefore P\left( A\cap B \right) =\frac { _{  }^{ 5 }{ { C }_{ 1 } } }{ ^{ 100 }{ { C }_{ 2 } } } $$ and $$\displaystyle P\left( A \right) =\frac { _{  }^{ 10 }{ { C }_{ 2 } } }{ ^{ 100 }{ { C }_{ 2 } } } $$
Then $$\displaystyle P\left( \frac { B }{ A }  \right) =\frac { P\left( A\cap B \right)  }{ P\left( A \right)  } =\frac { _{  }^{ 5 }{ { C }_{ 1 } } }{ ^{ 10 }{ { C }_{ 2 } } } =\frac { 1 }{ 9 } $$

Let, $${ E }_{ 1 }$$ be the event noted number is $$7$$.
$${ E }_{ 2 }$$ be the event noted number is $$8$$.
$$H$$ be getting head on coin.
$$T$$ be getting tail on coin.

Probability of choosing the first box $$= \dfrac { 3 }{ 6 } =\dfrac { 1 }{ 2 } $$
Probability of choosing the second box $$ = \dfrac { 3 }{ 6 } =\dfrac { 1 }{ 2 } $$
Probability of choosing a black ball
$$= $$(Probability of choosing the first bag * Probability of choosing black)
$$+$$ (Probability of choosing the second bag * Probability of choosing black)
 $$= \left (\dfrac { 1 }{ 2 } \times \dfrac { 3 }{ 7 } \right)+ \left(\dfrac { 1 }{ 2 } \times \dfrac { 4 }{ 7 } \right)$$
$$=\dfrac { 1 }{ 2 }$$
Now,
$$P\left(\dfrac { ball\quad chosen\quad from\quad first\quad box }{ chosen\quad ball\quad is\quad black } \right)$$

$$=\dfrac { P((ball\quad chosen\quad from\quad first\quad box)\bigcap  (chosen\quad ball\quad is\quad black)) }{ P(chosen\quad ball\quad is\quad black) } $$
Thus, required probability = $$\dfrac { \left (\dfrac { 1 }{ 2 } \times \dfrac { 3 }{ 7 } \right) }{ \dfrac { 1 }{ 2 }  } =\dfrac { 3 }{ 7 } $$

$$Statement 1:$$ If $$P\left( { H }_{ i }\cap E \right) =0$$ for some $$i$$, then
$$\displaystyle P\left( \frac { { H }_{ i } }{ E }  \right) =P\left( \frac { E }{ { H }_{ i } }  \right) =0$$
If $$P\left( { H }_{ i }\cap E \right)\neq 0 $$ for $$\forall i=1,2,...,n$$, then
$$\displaystyle P\left( \frac { { H }_{ i } }{ E }  \right) =\frac { P\left( { H }_{ i }\cap E \right)  }{ P\left( { H }_{ i } \right)  } \times \frac { P\left( { H }_{ i } \right)  }{ P\left( E \right)  } $$
$$\displaystyle =\frac { P\left( \frac { E }{ { H }_{ i } }  \right) \times P\left( { H }_{ i } \right)  }{ P\left( E \right)  } >P\left( \frac { E }{ { H }_{ i } }  \right) .P\left( { H }_{ i } \right) $$  $$\left[ \because 0<P\left( E \right) <1 \right] $$
Hence, statement $$1$$ may not always be true.
$$Statement 2:$$ Clearly $${ H }_{ 1 }\cup { H }_{ 2 }\cup ...\cup { H }_{ n }=S$$   (Sample Space)
$$\Rightarrow P\left( { H }_{ 1 } \right) +P\left( { H }_{ 2 } \right) +...+P\left( { H }_{ n } \right) =1$$

Let $$A$$ be the event that face 1 turns up
$$ B$$ be the event that face 2 turns up.
Then $$\displaystyle P\left ( A \right )= 0.10$$ and $$\displaystyle P\left ( B \right )= 0.32,$$
Since A, B are mutually exclusive, we have
$$\displaystyle P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )= 0.10+0.32= 0.42.$$
We are to find $$\displaystyle P\left ( A|A\cup B \right )$$
But $$\displaystyle P\left ( A|A\cup B \right )= \frac{P\left [ A\cap \left ( A\cup B \right ) \right ]}{P\left ( A\cup B \right )}$$

$$= \dfrac{P\left ( A \right )}{P\left ( A\cup B \right )}$$

Hence $$\displaystyle P\left ( A|A\cup B \right )= \frac{0.10}{0.42}= \frac{5}{21}$$

Total number of bags $$=5$$
Let A be the group in which there are $$3$$ bags
and B be  the group in which there are $$2$$ bags

Now, probability of selecting bag from group A is $$\dfrac{3}{5}$$
probability of selecting bag from group B is $$\dfrac{2}{5}$$

Now, probability of selecting a black ball from first group is $$\dfrac{3}{5} \times \dfrac{2}{7}=\dfrac{6}{35}$$
Also, probability of selecting a black ball from second group is $$\dfrac{2}{5} \times \dfrac{4}{5}=\dfrac{8}{25}$$

Probability of first group , given it is black is $$=\dfrac{\dfrac{6}{35}}{\dfrac{6}{35}+\dfrac{8}{25}}=\dfrac{15}{43}$$

Let  $$\displaystyle E_{1}$$ be the event that a subject is selected from first group.
$$\displaystyle E_{2}$$ the event that a subject is selected from the second group.
$$E$$ be the event that an engineering subject is selected.
Now the probability that die shows $$3$$ or $$5$$  is
$$\displaystyle P(E_1)=\frac{2}{6}=\frac{1}{3}$$