\(P(0)=\frac { 125 }{ 216 } ,P(1)=\frac { 75 }{ 216 } ,P(2)=\frac { 15 }{ 216 } ,P(3)=\frac { 1 }{ 216 } \)

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

Let 'p' be the probability of success and 'q' be tha probability of failure.
Then p + q = 1 and p = 3q
Solving \(p=\frac { 3 }{ 4 } ,q=\frac { 1 }{ 4 } \) and n = 5
Required probability = \(P(X\ge 3)=P(3)+P(4)+P(5)\)
= \(^{ 5 }{ C }_{ 3 }{ q }^{ 2 }{ p }^{ 3 }+^{ 5 }{ C }_{ 4 }{ q }^{ 1 }{ p }^{ 4 }+^{ 5 }{ C }_{ 5 }{ q }^{ 0 }{ p }^{ 5 }\)
= \(10{ \left( \frac { 1 }{ 4 } \right) }^{ 2 }{ \left( \frac { 3 }{ 4 } \right) }^{ 3 }+5{ \left( \frac { 1 }{ 4 } \right) }{ \left( \frac { 3 }{ 4 } \right) }^{ 4 }+\left( 1 \right) \left( 1 \right) { \left( \frac { 3 }{ 4 } \right) }^{ 5 }\)
\(={ \left( \frac { 3 }{ 4 } \right) }^{ 3 }\left[ \frac { 10 }{ 16 } +\frac { 15 }{ 16 } +\frac { 9 }{ 16 } \right] =\frac { 34 }{ 16 } { \left( \frac { 3 }{ 4 } \right) }^{ 3 }\)

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, \(P\left(\frac{C}{\bar{E}}\right)=\frac{P(C) \cdot P\left(\frac{\bar{E}}{C}\right)}{P(A) \times P\left(\frac{\bar{E}}{A}\right)+P(B) \times P\left(\frac{\bar{E}}{B}\right)+P(C) \times P\left(\frac{\bar{E}}{C}\right)}\)
\(\begin{aligned} =\frac{\frac{4}{7} \times 0.7}{\frac{1}{7} \times 0.2+\frac{2}{7} \times 0.5+\frac{4}{7} \times 0.7} \end{aligned}\)
\(\begin{aligned} =\frac{2.8 \times 7}{(0.2+1.0+2.8) \times 7}=\frac{2.8}{4}=0.7 \end{aligned}\)

Let the events be:
E₁ : Ball drawn from Bag I
E₂ : Ball drawn from Bag II and
A : Ball drawn is white
We have to find P(E₁/A)
Since both the bags are eaually likely to be selected,
\(P({ E }_{ 1 })=P({ E }_{ 2 })=\frac { 1 }{ 2 } \)
Also \(P(A/{ E }_{ 1 })=\frac { 4 }{ 7 } ,P(A/{ E }_{ 2 })=\frac { 3 }{ 10 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 1 }{ 2 } \right) \left( \frac { 4 }{ 7 } \right) }{ \left( \frac { 1 }{ 2 } \right) \left( \frac { 4 }{ 7 } \right) +\left( \frac { 1 }{ 2 } \right) \left( \frac { 3 }{ 10 } \right) } =\frac { \frac { 2 }{ 7 } }{ \frac { 2 }{ 7 } +\frac { 3 }{ 20 } } \)
\(=\frac { 2 }{ 7 } \times \frac { 140 }{ 40+21 } =\frac { 40 }{ 61 } \)  

Total number of outcomes = 36
For A, favourable outcomes are:
{(1, 6), (6, 1), (5, 2), (3, 4), (4, 3)}
P(A wins) = P(A) = \(\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
P(A loses) = \(P(\overset { \_ }{ A } )=1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)
For B, favourable outcomes are {(4, 6), (6, 4), (5, 5)}
P(B wins) = P(B) = \(\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
P(B loese) = \(P(\overset { \_ }{ B } )=1-\frac { 1 }{ 12 } =\frac { 11 }{ 12 } \)
Required probability
= \(P(\overset { \_ }{ A } B)+P(\overset { \_ }{ A } \overset { \_ }{ B } \overset { \_ }{ A } )+P(\overset { \_ }{ A } \overset { \_ }{ B } \overset { \_ }{ A } \overset { \_ }{ B } \overset { \_ }{ A } B)+.......\)   
\(P(\overset { \_ }{ A } )P(B)+P(\overset { \_ }{ A) } \overset { \_ }{ P(B } )P(\overset { \_ }{ A } )+P(\overset { \_ }{ A } )P(\overset { \_ }{ B } )P(\overset { \_ }{ A } )P(\overset { \_ }{ B } )P(\overset { \_ }{ A } )P(B)+.......\)
\(=​​\frac { 5 }{ 6 } \times \frac { 1 }{ 12 } +\frac { 5 }{ 6 } \times \frac { 11 }{ 12 } +\frac { 5 }{ 6 } \times \frac { 1 }{ 12 } +\frac { 5 }{ 6 } \times \frac { 11 }{ 12 } +\frac { 5 }{ 6 } \times \frac { 11 }{ 12 } +\frac { 5 }{ 6 } \times \frac { 1 }{ 12 } +......|G.P\)
\(=\frac { \frac { 5 }{ 6 } \times \frac { 1 }{ 12 } }{ 1-\frac { 5 }{ 6 } \times \frac { 11 }{ 12 } } =\frac { 5 }{ 17 } \)  

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

Here \(P(E\cap F)=\frac { 1 }{ 5 } \)
P(E)P(F) = \(\frac { 3 }{ 5 } \times \frac { 3 }{ 10 } =\frac { 9 }{ 50 } \)
Thus \(P(E\cap F)\neq P(E)P(F)\)
Hence, E and F are not independent.

Part (C) is the correct answer
Reason:\(P(A/B)=\frac { P(A\cap B) }{ P(B) } \)
\(=\frac { P(A\cap B) }{ 0 } \)
which is not defined

We have

LET E, D,T and M denote Easy, Diffcult,True/False and Multiple choice question respectively.
Total number of questions = 1400
Number of easy multiple choice questions = 500
\(\therefore \)\(P(E\cap M)=\frac { 500 }{ 1400 } \)
Total number of multiple choice questions = 900
\(\therefore \)\(P(M)=\frac { 900 }{ 1400 } \)
\(\therefore \)\(P(E/M)=\frac { P(E\cap M) }{ P(M) } \)
\(=\frac { \frac { 500 }{ 1400 } }{ \frac { 900 }{ 1400 } } =\frac { 5 }{ 9 } \)

Let A be the event that the machine produces 2 acceptable items.
Also let B₁ represent the event of correct set up and B₂ represent the event of incorrect setup.
Now \(\mathrm{P}\left(\mathrm{B}_1\right) =0.8, \mathrm{P}\left(\mathrm{B}_2\right)=0.2 \)
P(A|B₁ ) = 0.9 × 0.9 and P(A|B₂ ) = 0.4 × 0.4
Therefore
\({\mathrm{P}\left(\mathrm{B}_1 \mid \mathrm{A}\right)}=\frac{\mathrm{P}\left(\mathrm{B}_1\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{B}_1\right)}{\mathrm{P}\left(\mathrm{B}_1\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{B}_1\right) +\mathrm{P}\left(\mathrm{B}_2\right) \mathrm{P}\left({\left.\mathrm{A} \mid \mathrm{B}_2\right)}\right.}\)
\(=\frac{0.8 \times 0.9 \times 0.9}{0.8 \times 0.9 \times 0.9+0.2 \times 0.4 \times 0.4}=\frac{648}{680}=0.95 \)

(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 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)
= \(\frac { 4 }{ 52 } \times \frac { 3 }{ 51 } \times \frac { 4 }{ 50 } =\frac { 2 }{ 5525 } \)

Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F) or P (EF).
Now P(E) = P (black ball in first draw)\(=\frac{10}{15}\)
Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.
\(\text { i.e. }P(F \mid E)=\frac{9}{14} \)
By multiplication rule of probability, we have
P (E ∩ F) = P(E) P(F|E)
\(=\frac{10}{15} \times \frac{9}{14}=\frac{3}{7}\)
Multiplication rule of probability for more than two events If E, F and G are three events of sample space, we have
P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)
Similarly, the multiplication rule of probability can be extended for four or more events.
The following example illustrates the extension of multiplication rule of probability for three events.