(i) Total number of balls = 3 + 6 = 9
Let 'X' be the random variable when the red ball is drawn
Here 'X' takes values 0, 1, 2, 3, 4
P(X = 0) = P(No red ball)
= \(\frac { 3 }{ 9 } \times \frac { 3 }{ 9 } \times \frac { 3 }{ 9 } \times \frac { 3 }{ 9 } =\frac { 1 }{ 3 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 3 } =\frac { 1 }{ 81 } \)
P(X = 1) = P(one red ball)
= \(4\left( \frac { 6 }{ 9 } \times \frac { 3 }{ 9 } \times \frac { 3 }{ 9 } \times \frac { 3 }{ 9 } \right) =4\left( \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 3 } \right) =\frac { 8 }{ 81 } \)
P(X = 2) = P(Two red balls)
= \(6\left( \frac { 6 }{ 9 } \times \frac { 6 }{ 9 } \times \frac { 3 }{ 9 } \times \frac { 3 }{ 9 } \right) =6\left( \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 3 } \right) =\frac { 24 }{ 81 } \)
P(X = 3) = P(Tree red balls)
= \(4\left( \frac { 6 }{ 9 } \times \frac { 6 }{ 9 } \times \frac { 6 }{ 9 } \times \frac { 3 }{ 9 } \right) =6\left( \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \right) =\frac { 32 }{ 81 } \)
P(X = 4) = P(Four red balls)
= \(\left( \frac { 6 }{ 9 } \times \frac { 6 }{ 9 } \times \frac { 6 }{ 9 } \times \frac { 6 }{ 9 } \right) =\left( \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \right) =\frac { 16 }{ 81 } \)
Hence, the probability distribution is:     

(ii) Mean \(\sum { XP(X)=0 } \times \frac { 1 }{ 81 } +1\times \frac { 8 }{ 81 } +2\times \frac { 24 }{ 81 } +3\times \frac { 32 }{ 81 } +4\times \frac { 16 }{ 81 } \)
\(=0+\frac { 8 }{ 81 } +\frac { 48 }{ 81 } +\frac { 96 }{ 81 } +\frac { 64 }{ 81 } =\frac { 216 }{ 81 } =\frac { 8 }{ 3 } \)
Variance = \(\sum { { X }^{ 2 }P(X)-{ Mean }^{ 2 } } \)
\(=0\times \frac { 1 }{ 81 } +1\times \frac { 8 }{ 81 } +4\times \frac { 24 }{ 81 } +9\times \frac { 32 }{ 81 } +16\times \frac { 16 }{ 81 } -{ \left( \frac { 8 }{ 3 } \right) }^{ 2 }\)
\(=\frac { 648 }{ 81 } -\frac { 64 }{ 9 } =\frac { 8 }{ 9 } \)

Since A and B are independent evants,
\(\therefore \) \(\overset { \_ }{ A } \) and B, A and \(\overset { \_ }{ B } \) are also independent events 
Now \( P(\overset { \_ }{ A } \cap B)=\frac { 2 }{ 15 }\)  
\(\Rightarrow P(\overset { \_ }{ A } )P(B)=\frac { 2 }{ 15 } \)
\(\Rightarrow (1-P(A))P(B)=\frac { 2 }{ 15 } \)
\(\Rightarrow P(A)-P(A)P(B)=\frac { 2 }{ 15 } \).....(1)
And \(P(A\cap \overset { \_ }{ B) } =\frac { 1 }{ 6 } \)  \(\Rightarrow P(A)P(\overset { \_ }{ B } )=\frac { 1 }{ 6 } \)
\(\Rightarrow P(A)(1-P(B))=\frac { 1 }{ 6 } \)
\(\Rightarrow P(A)-P(A)P(B)=\frac { 1 }{ 6 } \).....(2)
PuttingP(A) = x and P(B) = y in(1) and (2), we get
\(y-xy=\frac { 2 }{ 15 } \)...(1)^'
and  \(x-xy=\frac { 1 }{ 6 } \)...(2)^'
Subtracting (2) find (1)
\(y-x=\frac { 2 }{ 15 } -\frac { 1 }{ 6 } \Rightarrow y-x=-\frac { 1 }{ 30 } \)
\(y=x-\frac { 1 }{ 30 } \)
Putting in (1) ,\(x-\frac { 1 }{ 30 } -x\left( x-\frac { 1 }{ 30 } \right) =\frac { 2 }{ 15 } \)
\(\Rightarrow x-\frac { 1 }{ 30 } -{ x }^{ 2 }+\frac { x }{ 30 } =\frac { 2 }{ 15 } \)
\(\Rightarrow 30x-1-30{ x }^{ 2 }+x=4\Rightarrow 30{ x }^{ 2 }-31x+5=0\)
Solving, \(x=\frac { 31\pm \sqrt { 961-600 } }{ 60 } =\frac { 31\pm 19 }{ 60 } =\frac { 5 }{ 6 } ,\frac { 1 }{ 5 } \)
When \(x=\frac { 5 }{ 6 } \), then from(3), \(y=\frac { 5 }{ 6 } -\frac { 1 }{ 30 } =\frac { 24 }{ 30 } =\frac { 4 }{ 5 } \)
when \(x=\frac { 1 }{ 5 } \),then from(3),\(y=\frac { 1 }{ 5 } -\frac { 1 }{ 30 } =\frac { 5 }{ 30 } =\frac { 1 }{ 6 } \)
Hence, \(P(A)=\frac { 5 }{ 6 } ,P(B)=\frac { 4 }{ 5 } orP(A)=\frac { 1 }{ 5 } ,P(B)=\frac { 1 }{ 6 } \)

P(not A and not B)
\(=P(\overset { - }{ A } \cap \overset { - }{ B } )=P(A\cup B)\)
\(\therefore \)  \(P(\overset { - }{ A } \cap \overset { - }{ B } )=1-P(A\cup B)\)
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(=\frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 8 } \)
\(=\frac { 2+4-1 }{ 8 } =\frac { 5 }{ 8 } \)
\(P(A\cap B)=1-\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \)

Let B₁, B₂ and G₁, G₂ be first, second boy and first, second girl respectively.
\(\therefore \) Sample space, S = {(G₁, G₂), (G₁, B₂), (B₁, G₂), (B₁, B₂)}.
Let E: Both children are girls = {(G₁, G₂)}
F: Youngest child is a girl = {(G₁, G₂), (B₁, G₂)} and G: at least one is a girl = {(G₁, G₂), (G₁,B₂), (B₁, G₂)}.
\(\therefore E\cap F\)= {(G₁, G₂), \(E\cap G\)= {G₁, G_2}.
\(\therefore P(E\cap F)\)= \(\frac { 1 }{ 4 } , P(E\cap G​​)=\frac { 1 }{ 4 } \), P(G) = \(\frac { 2 }{ 4 }\), P(G) = \(\frac { 3 }{ 4 }\).
(i)  P(E/F) = \(\frac { ​​P(E\cap F) }{ P(F) } =\frac { \frac { 1 }{ 4 } }{ \frac { 2 }{ 4 } } =\frac { 1 }{ 2 } \).
(ii) P(E/G) = \(\frac { ​​P(E\cap G) }{ P(G) } =\frac { \frac { 1 }{ 4 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 3 } \).

\(2 P(A)=P(B)=\frac{5}{13} \text { and } P(A \mid B)=\frac{2}{5} \)
\(\therefore \quad P(A)=\frac{5}{26}, P(B)=\frac{5}{13} \)
\(P(A \cap B)=P(A \mid B) \cdot P(B) \)
\(=\frac{2}{5} \times \frac{5}{13}=\frac{2}{13} \)
\(\text { Now } P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{11}{26} \)

We have
P(at least one of A and B) = P(A ∪ B)
= P(A) + P(B) − P(A ∩ B)
= P(A) + P(B) − P(A) P(B)
= P(A) + P(B) [1−P(A)]
= P(A) + P(B). P(A′)
= 1− P(A′) + P(B) P(A′)
= 1− P(A′) [1− P(B)]
= 1− P(A′) P (B′)

The sample space has 216 outcomes.
\(\text { Now } \quad \mathrm{A}=\left\{\begin{array}{lllll} (1,1,4) & (1,2,4) & \ldots & (1,6,4) & (2,1,4) & (2,2,4) & \ldots (2,6,4) \\ (3,1,4) & (3,2,4) & \ldots &(3,6,4) & (4,1,4) & (4,2,4) & \ldots(4,6,4) \\ (5,1,4) & (5,2,4) & \ldots & (5,6,4) & (6,1,4) & (6,2,4) & \ldots(6,6,4) \end{array}\right\}\)
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} and A ∩ B = {(6,5,4)}.
\(\text { Now }P(B)=\frac{6}{216} \text { and } P(A \cap B)=\frac{1}{216} \)
Then  \(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 1 }{ 216 } }{ \frac { 6 }{ 216 } } =\frac { 1 }{ 6 }\)

Let the events Be:
E₁ : coin is two headed
E₂ : coin is biased(heads 75%)
E₃ : Coin is biased(tails 40%)
and A : coin shows up head
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P({ E }_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=1,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } =\frac { 3 }{ 4 } ,\)
By Bayes' theorem
\(P({ E }_{ 1 }​/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } \times 1 }{ \frac { 1 }{ 3 } \times 1+\frac { 1 }{ 3 } \times \frac { 3 }{ 4 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 5 } } \)
\(=\frac{4}{4+3+2}=\frac{4}{9}\)

There are four entries determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1. Therefore, number of determinants that can be formed
= 2⁴ = 16
The value of determinant is positive in the following cases
\(\begin{vmatrix} 1 &0 \\0 &1 \end{vmatrix},\begin{vmatrix}1 &0 \\1 &1 \end{vmatrix},\begin{vmatrix}1 & 1 \\ 0 & 1 \end{vmatrix}\)
i.e, 3 determinants
Thus, the probability that the determinants is positive \(={3\over 16}\)

\(P(\bar { A } /\bar { B } )=\frac { P(\bar { A } \cap \bar { B } ) }{ P(\bar { B } ) } \)
\(=\frac { 1-P(A\cup B) }{ 1-P(B) } \)
\(=\frac { 1-[P(A)+P(B)-P(A\cap B)] }{ 1-P(B) } \)
\(=\frac { 7 }{ 10 } \)

Total number of bolts = 50
Total number of nuts = 50
Total number of rusted bolts = 25
Total number of rusted nuts = 25
Total no. of items = 100
Total no. rusted items = 50
E = rusted item, F = Bolts
\(P(E\cup F)=P(F)+P(F)-P(E\cap F)\)
\(P(E)=\left( \frac { 25 }{ 100 } \times \frac { 25 }{ 100 } \right) \)
\(=\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } =\frac { 1 }{ 16 } \)
\(P(F)=\left( \frac { 50 }{ 100 } \times \frac { 50 }{ 100 } \right) =\frac { 1 }{ 4 } \)
\(P(E\cap F)=\left( \frac { 25 }{ 100 } \times \frac { 25 }{ 100 } \right) \)
\(=\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } =\frac { 1 }{ 16 } \)
\(P(E\cup F)=\frac { 1 }{ 16 } +\frac { 1 }{ 4 } -\frac { 1 }{ 16 } \)
\(=\frac { 1 }{ 4 } \)

If E and F are independent events, then \(P\left( E\cap F \right) =P(E)\times P(F)\)
Now \(P(E\cup \bar { F } )=P(E)+P(F)-P(E\cap F)\)
\(\Rightarrow P(E\cup F)=P(E)+P(F)-P(E)\times P(F)\)
\(\Rightarrow P(E\cup F)=\frac { 1 }{ 3 } +\frac { 1 }{ 4 } -\frac { 1 }{ 3 } \times \frac { 1 }{ 4 } \)
\(\Rightarrow P(E\cup F)=\frac { 1 }{ 3 } +\frac { 1 }{ 4 } -\frac { 1 }{ 12 } =\frac { 4+3-1 }{ 12 } \)
\(\Rightarrow P(E\cup F)=\frac { 6 }{ 12 } \)
\(\therefore P(E\cup F)=\frac { 1 }{ 2 } \)

Given, \(P(E)=\frac { 1 }{ 4 } \)
\(P(F)=\frac { 1 }{ 2 } \)
and \(P(E\cap F) =\frac { 1 }{ 8 } \)
(a) P(E or F) = (E\(\cup\)F) = P(E) + P(F) - P(E\(\cap\)F)
\(=\frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 8 } \)
\(=\frac { 2+4-1 }{ 8 } \)
\(=\frac { 5 }{ 8 } \)
(b) P(not E and not F) = \(P(\bar { E } \cap \bar { F } )=P\overline { (E\cup F) } \)
= 1 - P(E\(\cup\) F)
= 1 - \(\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \)

Total number of cards = 2 + 6 + 8 = 16
(a) Number of red balls = 2
\(\therefore\)  Required probability = \(\frac { 2 }{ 16 } =\frac { 1 }{ 8 } \)
(b) Number of black balls = 6
\(\therefore\)  Required probability = \(\frac { 6 }{ 16 } =\frac { 3 }{ 8 } \)
(c) number of green balls = 8
\(\therefore\)  Required probability = \(\frac { 8 }{ 16 } =\frac { 1 }{ 2 } \)
(d) Number of non-red balls = 14
 \(\therefore\) Required probability = \(\frac { 14 }{ 16 } \)

The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is,
$S=\left\{\left(1,H\right),\left(1,T\right),\left(2,H\right),\left(2,T\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,H\right),\left(4,T\right),\left(5,H\right),\left(5,T\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\right\}$
Let $A$ be the event that the coin shows a tail and $B$ be the event that at least one die shows $3$.
$\therefore A=\left\{\left(1,T\right),\left(2,T\right),\left(4,T\right),\left(5,T\right)\right\}$
$B=\left\{\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(6,3\right)\right\}$
$\Rightarrow A\cap B=\varphi$
$\therefore P\left(A\cap B\right)=0$
Then, P(B) = P({3, 1}) + P({3, 2}) + P({3, 3}) + P({3, 4}) + P({3, 5}) + P({3, 6}) + P({6, 3})
\( =\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36} \\ =\frac{7}{36} \)
Probability of the event that the coin shows a tail, given that at least one die shows $3$, is given by $P\left(A|B\right)$.
Therefore, \(P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0}{\frac{7}{36}}=0\)

For independent events,
P(E∩F) = P(E) ⋅ P(F)
\(\Rightarrow 0.35=0.60 \times P(F) \Rightarrow P(F)=\frac{7}{12}=0.58\)

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

1/3

P(A) ⋅ P(B) = \(1\over2\)⋅\(1\over3\) = \(1\over6\) = P(A∩B) 
Yes, the events are independent.

The variance X takes values 3, 4, 5 and 6 
\(P(X=3)=\frac { 1 }{ 20 } ;\ P(X=4)=\frac { 3 }{ 20 } ;\)
\(P(X=5)=\frac { 6 }{ 20 } ;P(X=6)=\frac { 10 }{ 20 } ;\)

\(Mean=\sum { XP(X)=\frac { 105 }{ 20 } =\frac { 21 }{ 4 } } \)
\(Variance=\sum { X^{ 2 }P(X)-\{ \sum { X } P(X)\} ^{ 2 }=\frac { 63 }{ 80 } } \)
Alternative Method :
We have first six positive integers as 1, 2, 3, 4, 5, 6.
But X can take values 3, 4, 5, 6
[\(\because\) 1 and 2 cannot be selected since among 3 numbers the greatest is required]
P(X) = P(3 and other numbers which are less than 3)
i.e., (1, 2, 3)
\(=\frac { 1 }{ { 6 }_{ { C }_{ 3 } } } =\frac { 1 }{ \frac { 6\times 5\times 4 }{ 3\times 2 } } =\frac { 1 }{ 20 } \)
P(X = 4) = P(4 and other numbers which are less than 4) i.e., (1, 2, 4), (2, 3, 4), (1, 3, 4)
\(\frac { 3 }{ { { 6 }_{ C } }_{ 3 } } =\frac { 1 }{ \frac { 6\times 5\times 4 }{ 3\times 2 } } =\frac { 3 }{ 20 } \)
P(X=5) = P(5 and other numbers which are less than 5) i.e., (1, 2, 5), (1, 3,5), (1, 4, 5), (2, 3, 5), (2, 4, 5), (3, 4, 5).
\(\frac { 6 }{ { 6 }_{ { C }_{ 3 } } } =\frac { 6 }{ 20 } \)
P(X = 5) = P(6 and other numbers which are less than 6) i.e., [\(\because\) (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (2, 3, 6), (2, 4, 6), (2, 5, 6), (3, 4, 6), (3, 5, 6), (4, 5, 6)]
\(=\frac { 10 }{ { 6 }_{ { C }_{ 3 } } } =\frac { 10 }{ 20 } \)
\(\therefore\) Probability Distribution of X is

\(Mean\quad X=\sum { XP(X) } \)
\(=3\times \frac { 1 }{ 20 } +4\times \frac { 3 }{ 20 } +5\times \frac { 6 }{ 20 } +6\times \frac { 10 }{ 20 } \)
\(=\frac { 3 }{ 20 } +\frac { 12 }{ 20 } +\frac { 30 }{ 20 } +\frac { 60 }{ 20 } \)
\(=\frac { 105 }{ 20 } =5.25\)
\(Variance=\sum { X^{ 2 }P(X)-({ Mea }n)^{ 2 } } \)
\(={ (3) }^{ 2 }\times \frac { 1 }{ 20 } +{ (4) }^{ 2 }\times \frac { 3 }{ 20 } +{ (5) }^{ 2 }\times \frac { 6 }{ 20 } +{ (6) }^{ 2 }\times \frac { 10 }{ 20 } -{ \left( \frac { 105 }{ 20 } \right) }^{ 2 }\)
\(=9\times \frac { 1 }{ 20 } +16\times \frac { 3 }{ 20 } +25\times \frac { 6 }{ 20 } +36\times \frac { 10 }{ 20 } -\frac { 11025 }{ 400 } \)
\(=\frac { 9 }{ 20 } +\frac { 48 }{ 20 } +\frac { 150 }{ 20 } +\frac { 360 }{ 20 } -\frac { 11025 }{ 400 } \)
\(=\frac { 567 }{ 20 } -\frac { 11025 }{ 400 } \)
\(=\frac { 315 }{ 400 } \)
= 0.7875

Let the events be:
E₁ = Choosing 1^st coin
E₂ = Choosing 2^nd coin
E₃ = Choosing 3^rd coin
A: Getting Heads
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P(E_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=\frac { 40 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 1 }{ 2 } \)
\(P({ E }_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .\frac { 40 }{ 100 } }{ \frac { 1 }{ 3 } .\frac { 40 }{ 100 } +\frac { 1 }{ 3 } .\frac { 75 }{ 100 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } } =\frac { 8 }{ 33 } \)

Let, E₁: Event that lost card is a spade
E₂: Event that lost card is a not spade
A: Event that three spades are drawn without replacement from 51 cards
\(P({ E }_{ 1 })=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } ,\quad P({ E }_{ 2 })=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
\(P(A/{ E }_{ 1 })=\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } ,\quad P(A/{ E }_{ 2 })=\frac { 13_{ { C }_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } \)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } }{ \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } +\frac { 3 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } } \)
\(=\frac { 10 }{ 49 } \)