P(getting on odd number) = \(\frac { 3 }{ 6 } \)
= \(\frac { 1 }{ 2 } \) [\(\because \) 1, 3, 5 are odd]
\(\therefore \) P(getting an even number) = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
\(\therefore \) Probability of getting an odd number on throw of adie 3 times = Probability of getting an even number in a throw of a die 3 times
= \(\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
Hence,probability of at least one odd number in athrow of adie 3 times
= \(1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)

Since A and B are independent events
\(\therefore \) \(P(A\cap B)=P(A)P(B)\)
\(=\left( \frac { 3 }{ 5 } \right) \left( \frac { 1 }{ 5 } \right) =\frac { 3 }{ 25 } \)

(i) \(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { 0.2 }{ 0.3 } =\frac { 2 }{ 3 } \)
(ii) \(P(F/E)=\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.6 } =\frac { 1 }{ 3 }\)

We know that the sample space is S = {1, 2, 3, 4, 5, 6}
Now E = { 3, 6}, F = { 2, 4, 6} and E ∩ F = {6}
Then \(\mathrm{P}(\mathrm{E})=\frac{2}{6}=\frac{1}{3}, \mathrm{P}(\mathrm{F})=\frac{3}{6}=\frac{1}{2} \text { and } \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{6}\)
Clearly P(E ∩ F) = P(E). P (F)
Hence E and F are independent events.

Let K denote the event that the card drawn is king and A be the event that the card drawn is an ace.
Clearly, we have to find P (KKA)
Now   P(K) = \(\frac { 4 }{ 52 } \)
Also, P (K|K) is the probability of second king with the condition that one king has already been drawn. Now there are three kings in (52 − 1) = 51 cards.      
Therefore P(K/K) = \(\frac { 3 }{ 51 } \)
Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now there are four aces in left 50 cards.
Therefore P(A|KK) =\(\frac { 4 }{ 50 } \)
By multiplication law of probability, we have
Now   P(KKA) = P(K) P(K/K) P(A/KK)
= \(\frac { 4 }{ 52 } \times \frac { 3 }{ 51 } \times \frac { 4 }{ 50 } =\frac { 2 }{ 5525 } \)

Total cards are 12
A : number drawn is more than 3, i.e. 4, 5, 6, ..., 12.
B : getting an even number, i.e. 2, 4, 6, 8, 10, 12.
\(A \cap B: 4,6,8,10,12 \)
\(P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{5 / 12}{9 / 12}=\frac{5}{9} . \)

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 \(={3\over 16}\)

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

\(P(E\cap F)=P(E)\times P(F)\)
\(P(E\cup F)=P(E)+P(F)-P(E\cap F)\)
\(\Rightarrow P(E\cup F)=P(E)+P(F)-P(E)\times P(F)\)
\(\Rightarrow 0.85=P(E)+0.35-P(E)\times 0.35\)
\(\Rightarrow 0.85=P(E)(1-0.35)+0.35\)
\(\Rightarrow 0.85-0.35=P(E)(0.65)\)
\(\Rightarrow \frac { 0.52 }{ 0.65 } =P(E)\)
\(\Rightarrow P(E)=\frac { 50 }{ 65 } =\frac { 10 }{ 13 } \)
\(\therefore P(F)=1-\frac { 10 }{ 13 } =\frac { 3 }{ 13 } \)

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)
\(\Rightarrow 0.35=0.60 \times P(F) \Rightarrow P(F)=\frac{7}{12}=0.58\)

1/3

proof :
By formula of conditional probability, we know that
\( \mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right) =\frac{\mathrm{P}\left(\mathrm{A} \cap \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})} \)
\(=\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE} \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})}(b y \) (by multiplication rule of probability)
\( =\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE}_i\right)}{\sum_{j=1}^n \mathrm{P}\left(\mathrm{E}_j\right) \mathrm{P}\left(\mathrm{AlE}_j\right)} \)(by the result of theorem of total probability)

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

Total number of ways of selecting two numbers = ⁶C₂ = 15
Values of x (larger of the two) can be 2, 3, 4, 5, 6
\(P(x=2)={1\over 15}\)
\(P(x=3)={2\over 15}\)
\(P(x=4)={3\over 15}\)
\(P(x=5)={4\over 15}\)
and \(P(x=6)={5\over 15}\)
Distribution can be written as

Mean \(=\sum xP(x)={70\over 15}={14\over 3}\)

Let E₁: urn has 2 white balls
E₂: urn has 3 white balls
E₃: urn has 4 white balls
A: 2 balls drawn are white
P(E₁) = P(E₂) = P(E₃) = \(\frac { 1 }{ 3 } \)
P(A/E₁)=\(\frac { { 2 }_{ { C }_{ 2 } } }{ { 4 }_{ { C }_{ 2 } } } =\frac { 1 }{ 6, } \)
\(P(A/{ E }_{ 2 })=\frac { { 2 }_{ { C }_{ 2 } } }{ { 4 }_{ C_{ 2 } } } =\frac { 1 }{ 6 } ,\)
\(P(A/{ E }_{ 2 })=\frac { { 3 }_{ { C }_{ 2 } } }{ { 4 }_{ C_{ 2 } } } =\frac { 1 }{ 2 } ,\)
\(P(A/{ E }_{ 3 })=\frac { { 4 }_{ { C }_{ 2 } } }{ { 4 }_{ { C }_{ 2 } } } =1\)
\(\therefore P({ E }_{ 3 }/A)\)
\(=\frac { { P(E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .1 }{ \frac { 1 }{ 3 } .\frac { 1 }{ 6 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } +\frac { 1 }{ 3 } .1 } \)
\(=\frac { 6 }{ 10 } =0.6\)