Given n = 4
Getting  \(p=\frac { 1 }{ 2 } \ and\ q=\frac { 1 }{ 2 } \)

Mean = \(\sum { xp(x)=\frac { 32 }{ 16 } =2 } \)
\(\\ Variance=\sum { { x }^{ 2 }p(x)-\{ \sum { P(x){ \} }^{ 2 } } } \)
\(=\frac { 80 }{ 16 } -{ (2) }^{ 2 }=5-4=1\)

Let us define the following events
E₁ : Bag X is selected
E₂ : Bag Y is selected
and E : Getting one white and one black ball in a draw of two balls.
Here, \(P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}\)
[\(\because\) probability of selecting each bag is equal]
Now, \(P\left(\frac{E}{E_1}\right)\) = Probability of drawing one white and one black ball from bag X
\(=\frac{{ }^4 C_1 \times{ }^2 C_1}{{ }^6 C_2}=\frac{4 \times 2}{\frac{6 \times 5}{2 \times 1}}=\frac{16}{6 \times 5}=\frac{8}{15}\)
and \(P\left(\frac{E}{E_2}\right)=\) Probability of drawing one white and one black ball from bag Y
\(=\frac{{ }^3 C_1 \times{ }^3 C_1}{{ }^6 C_2}=\frac{3 \times 3}{\frac{6 \times 5}{2 \times 1}}=\frac{3}{5}\)
\(\therefore\) The probability that the one white and one black balls are drawn from bag Y,
\(P\left(\frac{E_2}{E}\right)=\frac{P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)}\)
                                 [by using Baye's theorem]
\(=\frac{\frac{1}{2} \times \frac{3}{5}}{\frac{1}{2} \times \frac{8}{15}+\frac{1}{2} \times \frac{3}{5}}=\frac{\frac{3}{5}}{\frac{8}{15}+\frac{3}{5}}=\frac{\frac{3}{5}}{\frac{8+9}{15}}=\frac{3 \times 15}{5 \times 17}=\frac{9}{17}\)

Let the events be as below:
E₁ : Plant 1 is chosen
E₂ : Plant 2 is chosen
and A : Motorcycle is of standard quality
We have :\(P({ E }_{ 1 })=\frac { 70 }{ 100 } P({ E }_{ 2 })=\frac { 30 }{ 100 } \)
\(P({ A/E }_{ 1 })\frac { 80 }{ 100 } P(A/{ E }_{ 2 })=\frac { 90 }{ 100 } \)
By Bayes' Theorem,
(i) \(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 70 }{ 100 } \right) \left( \frac { 80 }{ 100 } \right) }{ \left( \frac { 70 }{ 100 } \right) \left( \frac { 80 }{ 100 } \right) +\left( \frac { 30 }{ 100 } \right) \left( \frac { 90 }{ 100 } \right) } \)
\(=\frac { 5600 }{ 5600+2700 } =\frac { 56 }{ 83 } \)
(ii) \(P({ E }_{ 2 }/A)=\frac { P({ E }_{ 2 })P(A/{ E }_{ 21 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 30 }{ 100 } \right) \left( \frac { 90 }{ 100 } \right) }{ \left( \frac { 70 }{ 100 } \right) \left( \frac { 80 }{ 100 } \right) +\left( \frac { 30 }{ 100 } \right) \left( \frac { 90 }{ 100 } \right) } \)
\(=\frac { 2700 }{ 2700+5600 } =\frac { 27 }{ 83 } \)
Riding a motorcycle involves higher risk but children ride at very high speed and do stunts causing accidents.

Let us define the following events
A = Selecting person A
B = Selecting person B
C = Selecting person C
\(P(A)=\frac{1}{1+2+4}, P(B)=\frac{2}{1+2+4}\)
and \(P(C)=\frac{4}{1+2+4}\)
\(\Rightarrow \quad P(A)=\frac{1}{7}, P(B)=\frac{2}{7} \text { and } P(C)=\frac{4}{7}\)
Let E = Event to introduce the changes in their profit.
Also, given \(P\left(\frac{E}{A}\right)=0.8, P\left(\frac{E}{B}\right)=0.5 \text { and } P\left(\frac{E}{C}\right)=0.3\)
\(\begin{aligned} \Rightarrow P\left(\frac{\bar{E}}{A}\right)=1-0.8=0.2, P\left(\frac{\bar{E}}{B}\right)=1-0.5=0.5 \end{aligned}\)
and \(\begin{aligned} \text { and } P\left(\frac{\bar{E}}{C}\right)=1-0.3=0.7 \end{aligned}\)
The probability that change does not take place by the appointment of C, \(P\left(\frac{C}{\bar{E}}\right)=\frac{P(C) \cdot P\left(\frac{\bar{E}}{C}\right)}{P(A) \times P\left(\frac{\bar{E}}{A}\right)+P(B) \times P\left(\frac{\bar{E}}{B}\right)+P(C) \times P\left(\frac{\bar{E}}{C}\right)}\)
\(\begin{aligned} =\frac{\frac{4}{7} \times 0.7}{\frac{1}{7} \times 0.2+\frac{2}{7} \times 0.5+\frac{4}{7} \times 0.7} \end{aligned}\)
\(\begin{aligned} =\frac{2.8 \times 7}{(0.2+1.0+2.8) \times 7}=\frac{2.8}{4}=0.7 \end{aligned}\)

Here sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
where H = Head and T = Tail.
Here X takes value 0, 1, 2 and 3
\(P(0)=\frac { 1 }{ 8 } \)   [{TTT}]
\(P(1)=\frac { 3 }{ 8 } \)   [{HTT}, {THT}, {TTH}]
\(P(2)=\frac { 3 }{ 8 } \)  [{HHT}, {HTH}, {THH}]
\(P(3)=\frac { 1 }{ 8 } \) [{HHH}]
Probability distribution is:

\(E(X)=0\times \frac { 1 }{ 8 } +1\times \frac { 3 }{ 8 } +2\times \frac { 3 }{ 8 } +3\times \frac { 1 }{ 8 } \)
\(=0+\frac { 3 }{ 8 } +\frac { 6 }{ 8 } +\frac { 3 }{ 8 } =\frac { 12 }{ 8 } =\frac { 3 }{ 2 } =1.5\)

Part (A) is the correct answer
Let E₁ and E₂ be the events when speaks the truth or not respectively.
\(\therefore \)  \(P({ E }_{ 1 })=\frac { 4 }{ 5 } ,P({ E }_{ 2 })=1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Let A be the event when head appears.
\(\therefore \)  \(P(A/{ E }_{ 1 })=\frac { 1 }{ 2 } ,P(A/{ E }_{ 2 })=\frac { 1 }{ 2 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 4 }{ 5 } \right) \left( \frac { 1 }{ 2 } \right) }{ \left( \frac { 4 }{ 5 } \right) \left( \frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 5 } \right) \left( \frac { 1 }{ 2 } \right) } =\frac { \frac { 4 }{ 10 } }{ \frac { 1 }{ 2 } } =\frac { 4 }{ 5 } \)

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) = \(\frac { ​​P(E\cap G) }{ P(G) } =\frac { \frac { 1 }{ 4 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 3 } \).

(i) The number of trials is finite. When the drawing is done with replacement, the probability of success (say, red ball) is p = \(\frac{7}{16}\)which is same for all six trials (draws). Hence, the drawing of balls with replacements are Bernoulli trials.
(ii) When the drawing is done without replacement, the probability of success (i.e., red ball) in first trial is \(\frac{7}{16}\)in 2nd trial is \(\frac{6}{15}\) if the first ball drawn is red or \(\frac{7}{15}\) if the first ball drawn is black and so on. Clearly, the probability of success is not same for all trials, hence the trials are not Bernoulli trials

Let b stand for boy and g for girl. The sample space of the experiment is
S = {(b, b), (g, b), (b, g), (g, g)}
Let E and F denote the following events :
E : ‘both the children are boys’
F : ‘at least one of the child is a boy’
Then E = {(b, b)} and F = {(b, b), (g, b), (b, g)}
Now E ∩ F = {(b, b)}
Thus P(F) = \(\frac { 3 }{ 4 } \)and P (E ∩ F ) = \(\frac { 1 }{ 4 } \)
Therefore P(E/F)  \(=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\)

Let E₁ be the event of choosing the bag I, E₂ the event of choosing the bag II and A be the event of drawing a red ball.
Then P(E₁ ) = P(E₂ ) = 1/2
Also P(A|E₁ ) = P(drawing a red ball from Bag I) = 3/7
and P(A|E₂ ) = P(drawing a red ball from Bag II) = 5/11
Now, the probability of drawing a ball from Bag II, being given that it is red, is P(E₂ |A)
By using Bayes' theorem, we have
\(\mathrm{P}\left(\mathrm{E}_2 \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A|E}_2\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A|E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A|E}_2\right)}=\frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{5}{11}}=\frac{35}{68}\)

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 \(={3\over 16}\)

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

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

P(E\(\cap\)F) = P(E) + P(F) - P(E\(\cup\)F)
\(=\frac { 6 }{ 11 } +\frac { 5 }{ 11 } -\frac { 7 }{ 11 } \)
\(=\frac { 4 }{ 11 } \)
(a)  \(P(E/F)=\frac { P(E\cap F) }{ P(F) } \)
\(=\frac { \frac { 4 }{ 11 } }{ \frac { 5 }{ 11 } } =\frac { 4 }{ 5 } \)
(b) P(E/E) = \(\frac { P(E\cap F) }{ P(E) } \)
\(=\frac { \frac { 4 }{ 11 } }{ \frac { 6 }{ 11 } } =\frac { 4 }{ 6 } \)
\(=\frac { 2 }{ 3 } \)

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 = \(\frac { 14 }{ 16 } \)

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

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)

Let C₁, C₂, C₃, C₄ be the events that the lost card is of heart, spades, diamond or club respectively.
Obviously P(C₁) = P(C₂) = P(C₃) = P(C₄)
\(={13\over 52}={1\over 4}\)
Let S be the event of drawing two cards of heart from the remaining 51 cards.We wish to find \(P\left(C_1\over S\right)\)
Now \(P\left(C_1\over S\right)\) is the probability of drawing two heart cards from 51 cards given that one heart card is lost
\(={^{12}C_2\over ^{51}C_2}={12\times11\over 1\times2}\times{1\times2\over 51\times50}={22\over 425}\)
\(P\left( S\over C_3\right)=P\left( S\over C_3\right)=P\left( S\over C_4\right)={^{13}C_2\over ^{51}C_2}\)
\(={13\times12\over 1\times2}\times{1\times2\over 51\times50}={26\over 425}\)
By Bayes' Theorem
\(P\left(C_1\over S\right)={P(C_1).P\left(S\over C_1\right)\over \sum P(C_1).P\left(S\over C_1\right)}\)
\(={{1\over4}\times{22\over 425}\over{1\over 4}\times{22\over 425}+{1\over 4}\times{26\over 425}+{1\over4}\times{26\over 425}}\)
\(={22\over 22+26+26+26}\)
\(={11\over 50}\)

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