Probability

Here, five students S₁, S₁, S₃, S₄ and S₅ and five seats R₁, R₂, R₃, R₄ and R₅

$ \therefore $ Total number of arrangement of sitting five students is 5! = 120

Here, S₁ gets previously allotted seat R₁

$ \therefore $ S₂, S₃, S₄ and S₅ not get previously seats.

Total number of way S₂, S₃, S₄ and S₅ not get previously seats is

$4!\left( {1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}} \right)$

$ = 24\left( {1 - 1 + {1 \over 2} - {1 \over 6} + {1 \over {24}}} \right)$

$ = 24\left( {{{12 - 4 + 1} \over {24}}} \right) = 9$

$ \therefore $ Required probability = ${9 \over {120}} = {3 \over {40}}$

Here, $n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$ = Total $ - n(\overline {{T_1}} \cup \overline {{T_2}} \cup \overline {{T_3}} \cup \overline {{T_4}} )$

$ \Rightarrow $ $n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$

$ = 5! - [{}^4{C_1}4!2!\, - ({}^3{C_1}3!2! + {}^3{C_1}3!2!2!) + ({}^2{C_1}2!2!\, + \,{}^4{C_1}2.2!) - 2]$

$ \Rightarrow $ $n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$

$ = 120 - [192 - (36 + 72) + (8 + 16) - 2]$

$ = 120 - [192 - 108 + 24 - 2] = 14$

$ \therefore $ Required probability = ${{14} \over {120}} = {7 \over {60}}$

Let P(E) = x and P(F) = y

Now, $P(E \cap F) = {1 \over {12}}$

$ \Rightarrow P(E)P(F) = {1 \over {12}}$

$ \Rightarrow xy = {1 \over {12}}$

Also, $P(E' \cap F') = {1 \over 2}$

$ \Rightarrow (1 - P(E))(1 - P(F)) = {1 \over 2}$

$ \Rightarrow (1 - x)(1 - y) = {1 \over 2}$

$ \Rightarrow 1 - x - y + xy = {1 \over 2}$

$ \Rightarrow x + y = 1 + xy - {1 \over 2}$

$ \Rightarrow x + y = 1 + {1 \over {12}} - {1 \over 2}$

$ \Rightarrow x + y = {1 \over 2} + {1 \over {12}} = {7 \over {12}}$ ........ (1)

Now,

${(x - y)^2} = {(x + y)^2} - 4xy$

$ \Rightarrow {(x - y)^2} = {{49} \over {144}} - {1 \over 3} \Rightarrow {{49 - 98} \over {144}} \Rightarrow {1 \over {144}}$

$ \Rightarrow (x - y) = {1 \over {12}} \Rightarrow x - y = {1 \over 2}$ ........ (2)

From Eqs. (1) and (2), we get

$(x + y)(x - y) = {7 \over {12}} + {1 \over {12}}$

$ \Rightarrow x = {4 \over {12}};y = {3 \over {12}}$

$ \Rightarrow {x \over y} = {4 \over 3} = {{P(E)} \over {P(F)}}$

The number of ways to form a committee having at least one woman is

$ = {}^5{C_1} \times {}^{10}{C_3} + {}^5{C_2} \times {}^{10}{C_2} + {}^5{C_3} \times {}^{10}{C_1} + {}^5{C_4}$

$ = {{5!} \over {4!}} \times {{10!} \over {7! \times 3!}} + {{5!} \over {2!3!}} \times {{10!} \over {8!2!}} + {{5!} \over {3!2!}} \times {{10!} \over {9!1!}} + {{5!} \over {4!}}$

$ = 5 \times {{10 \times 9 \times 8} \over {3 \times 2}} + {{5 \times 4} \over {2 \times 1}} \times {{10 \times 9} \over 2} + {{5 \times 4} \over {2 \times 1}} \times 10 + 5$

$ = 600 + 450 + 100 + 5 = 1155$

The number of ways to form a committee having more women than men is

${}^5{C_2} \times {}^{10}{C_2} + {}^5{C_4} = {{5!} \over {6!2!}} \times {{10!} \over {9!}} + {{5!} \over {4!}}$

$ = 10 \times 10 + 5 = 105$

Therefore, the probability for the committees to have more women than men is

${{105} \over {1155}} = {1 \over {11}}$

An unbiased coin is tossed 8 times which is same as 8 coins tossed 1 times.

$\therefore\,\,\,$ Possible no. of out come = 2⁸

$\therefore\,\,\,$ Sample space = 2⁸

Here in this condition, all head or all tail out come is not acceptable.

No. of times all head can occur

(H H H H H H H H) = 1

$\therefore\,\,\,$ Probability (all head) = ${1 \over {{2^8}}}$ = ${1 \over {256}}$

No. of times all tail can occur

(T T T T T T T T) = 1

$\therefore\,\,\,$ Probability (all tail) = ${1 \over {{2^8}}}$ = ${{1 \over {256}}}$

$\therefore\,\,\,$ Required probability

= 1 $-$ (P (All head) + P (All tail))

= 1 $-$ ( ${{1 \over {256}}}$ + ${{1 \over {256}}}$)

= 1 $-$ ${{1 \over {128}}}$

= ${{127} \over {128}}$

We have the following probabilities:

$\bullet$ The probability that the target is hit by the person P is ${3 \over 4}$.

$\bullet$ The probability that the target is not hit by the person P is $1 - {3 \over 4} = {1 \over 4}$.

$\bullet$ The probability that the target is hit by the person Q is ${1 \over 2}$.

$\bullet$ The probability that the target is not hit by the person Q is $1 - {1 \over 2} = {1 \over 2}$.

$\bullet$ The probability that the target is hit by the person R is ${5 \over 8}$.

$\bullet$ The probability that the target is not hit by the person R is $1 - {5 \over 8} = {3 \over 8}$.

Here, we have used the fact that if the probability of occurrence of an event is p, then the probability of non-occurrence of an event is $q = 1 - p$.

Therefore, the probability that the target is hit by P or Q and not by R is

(Probability that the target is hit by P and not by Q and R) + (Probability that the target is hit by Q and not by P and R) + (Probability that the target is hit by both P and Q and not by R)

$ = \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{1 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right)$

$ = {9 \over {64}} + {3 \over {64}} + {9 \over {64}} = {{9 + 3 + 9} \over {64}} = {{21} \over {64}}$

We can apply binomial probability distribution

n = 10

p = Probability of drawing a green ball = ${{15} \over {25}}$ = ${3 \over 5}$

Also q = 1 - ${3 \over 5}$ = ${2 \over 5}$

Variance = npq

= $10 \times {3 \over 5} \times {2 \over 5}$ = ${{12} \over 5}$

Let A = {0, 1, 2, 3, 4, ......., 10}

Total number of ways of selecting 2 different numbers from A is

n (S) = ¹¹C₂ = 55, where 'S' denotes sample space

Let E be the given event

$ \therefore $ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}

$ \Rightarrow $ n (E) = 6

$ \therefore $ P(E) = ${{n\left( E \right)} \over {n\left( S \right)}}$ = ${6 \over {55}}$

Given, P (A $ \cap $ B $ \cap $ C) = ${1 \over {16}}$

P (exactly one of A or B occurs)

= P(A) + P (B) – 2P (A $ \cap $ B) = ${1 \over 4}$ .....(1)

P (Exactly one of B or C occurs)

= P(B) + P (C) – 2P (B $ \cap $ C) = ${1 \over 4}$ .....(2)

P (Exactly one of C or A occurs)

= P(C) + P(A) – 2P (C $ \cap $ A) = ${1 \over 4}$ .....(3)

Adding (1), (2) and (3),we get

2[ P(A) + P(B) + P (C) - P (A $ \cap $ B)

- P (B $ \cap $ C) - P (C $ \cap $ A)] = ${3 \over 4}$

$ \Rightarrow $ P(A) + P(B) + P (C) - P (A $ \cap $ B)

- P (B $ \cap $ C) - P (C $ \cap $ A) = ${3 \over 8}$

$ \therefore $ P(atleast one event occurs)

= P (A $ \cup $ B $ \cup $ C)

= P(A) + P(B) + P (C) - P (A $ \cap $ B)

- P (B $ \cap $ C) - P (C $ \cap $ A) + P (A $ \cap $ B $ \cap $ C)

= ${3 \over 8} + {1 \over {16}}$ = ${7 \over {16}}$

To solve this problem, we need to find the total number of possible solutions to the equation $ x + y + z = 10 $ where $ x, y, $ and $ z $ are nonnegative integers.

First, we employ the "stars and bars" method to determine the number of nonnegative integer solutions to this equation. This method states that the number of ways to partition $ n $ identical items (our sum) into $ k $ distinct groups (our variables) is given by:

$ \binom{n + k - 1}{k - 1} $

For $ x + y + z = 10 $, we have $ n = 10 $ and $ k = 3 $. Plugging into the formula, we get:

$ \binom{10 + 3 - 1}{3 - 1} = \binom{12}{2} $

We calculate the binomial coefficient:

$ \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 $

So, there are 66 possible solutions to the equation $ x + y + z = 10 $.

Next, we need to find the number of solutions for which $ z $ is even. Let $ z = 2k $ where $ k $ is a nonnegative integer. Then the equation becomes:

$ x + y + 2k = 10 $

Rearranging it, we get:

$ x + y = 10 - 2k $

Here, $ 10 - 2k $ must be nonnegative, so $ k $ can take values $ 0, 1, 2, 3, 4, 5 $ making total 6 possible values of k.

For each value of $ k $, $ x + y $ must equal the corresponding $ 10 - 2k $. The number of nonnegative integer solutions to $ x + y = m $ for any nonnegative integer $ m $ is given by:

$ \binom{m + 1}{1} = m + 1 $

We will sum the solutions for each valid $ k $:

When $ k = 0 $: $ x + y = 10 $, the number of solutions is $ 11 $.

When $ k = 1 $: $ x + y = 8 $, the number of solutions is $ 9 $.

When $ k = 2 $: $ x + y = 6 $, the number of solutions is $ 7 $.

When $ k = 3 $: $ x + y = 4 $, the number of solutions is $ 5 $.

When $ k = 4 $: $ x + y = 2 $, the number of solutions is $ 3 $.

When $ k = 5 $: $ x + y = 0 $, the number of solutions is $ 1 $.

Summing these, we get:

$ 11 + 9 + 7 + 5 + 3 + 1 = 36 $

Thus, there are 36 solutions where $ z $ is even out of a total of 66 solutions. Hence, the probability that $ z $ is even is given by:

$ \frac{36}{66} = \frac{6}{11} $

Thus, the correct answer is:

Option C: $ \frac{6}{11} $

$P(X) = {1 \over 3}$

$P\left( {{X \over Y}} \right) = {{P(X \cap Y)} \over {P(Y)}} = {1 \over 2}$

$P\left( {{Y \over X}} \right) = {{P(X \cap Y)} \over {P(X)}} = {2 \over 5}$

$P(X \cap Y) = {2 \over {15}}$

$P(Y) = {4 \over {15}}$

$P\left( {{{X'} \over Y}} \right) = {{P(Y) - P(X \cap Y)} \over {P(Y)}}$

$ = {{{4 \over {15}} - {2 \over {15}}} \over {{4 \over {15}}}} = {1 \over 2}$

$P(X \cup Y) = $${1 \over 3} + {4 \over {15}} - {2 \over {15}}$

$ = {7 \over {15}} = {7 \over {15}}$

Probability of fail, P(F) = p

$ \therefore $   Probability of success, P(S) = 2p

We know,

p + 2Pp = 1

$ \Rightarrow $   p = ${1 \over 3}$

$ \therefore $   Now, probability of at least 5 success in 6 trials,

P(x $ \ge $ 5)

= P(x = 5) + P (x = 6)

= ⁶C₅ ${\left( {{2 \over 3}} \right)^5}{\left( {{1 \over 3}} \right)^1}{ + ^6}{C_6}{\left( {{2 \over 3}} \right)^6}{\left( {{1 \over 3}} \right)^o}$

= ${\left( {{2 \over 3}} \right)^5}\left( {{6 \over 3} + {2 \over 3}} \right)$

= ${{256} \over {729}}$

$P\left( {{A \over {A' \cup B'}}} \right)$

$ = {{P\left[ {A \cap \left( {A' \cup B'} \right)} \right]} \over {P\left( {A' \cap B'} \right)}}$

$ = {{P\left[ {\left( {A \cap A'} \right) \cup \left( {A \cap B'} \right)} \right]} \over {P\left( {A \cap B} \right)'}}$

$\left[ \, \right.$As   $\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]$

$ = {{P\left( {A \cap B'} \right)} \over {P\left( {A \cap B} \right)'}}$

As   $A \cap A' = \phi $

$ \therefore $   $\phi \cup \left( {A \cap B'} \right) = A \cap B'$

$ = {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}$

$\left[ \, \right.$As   $A \cap B' = A - \left( {A \cap B} \right)$

and   $\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]$

Given  $P\left( A \right) = {2 \over 5}$

and   $P\left( {A \cap B} \right) = {3 \over {20}}$

$ = {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}$

$ = {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}$

$ = {5 \over {17}}$

Total possible outcome with two six faced dice = 6² = 36

When dice A shows up 4, the possible cases are
E₁ = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
$\therefore$ $P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}$

When dice B shows up 2, the possible cases are
E₂ = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases

$P\left( {{E_2}} \right) = {6 \over {36}} = {1 \over 6}$

${E_1} \cap {E_2}$ = Common in both in E₁ and E₂ = { (4, 2) }

$P\left( {{E_1} \cap {E_2}} \right) = {1 \over {36}}$

And $P\left( {{E_1}} \right)$.$P\left( {{E_2}} \right)$ = ${1 \over 6}$.${1 \over 6}$ = ${1 \over 36}$

$\therefore$$P\left( {{E_1} \cap {E_2}} \right)$ = $P\left( {{E_1}} \right)$.$P\left( {{E_2}} \right)$

$\therefore$ E₁ and E₂ are independent.

${E_3}$ = [ (1, 2), (1, 4), (1, 6),
           (2, 1), (2, 3), (2, 5),
           (3, 2), (3, 4), (3, 6),
           (4, 1), (4, 3), (4, 5),
           (5, 2), (5, 4), (5, 6),
           (6, 1), (6, 3), (6, 5) ] = 18 cases

${E_1} \cap {E_3}$ = { (4, 1) (4, 3) (4, 5) } = 3 cases

$\therefore$ $P\left( {{E_1} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$ = ${1 \over 6} \times {1 \over 2}$ = $P\left( {{E_1}} \right)$ $ \times $ $P\left( {{E_3}} \right)$

$\therefore$ E₁ and E₃ are independent.

${E_2} \cap {E_3}$ = { (1, 2) (3, 2) (5, 2) } = 3 cases

$\therefore$ $P\left( {{E_2} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$ = ${1 \over 6} \times {1 \over 2}$ = $P\left( {{E_2}} \right)$ $ \times $ $P\left( {{E_3}} \right)$

$\therefore$ E₂ and E₃ are independent.

${E_1} \cap {E_2} \cap {E_3}$ = 0

$\therefore$ $P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)$ = 0

$P\left( {{E_1}} \right) \times $ $P\left( {{E_2}} \right) \times $$P\left( {{E_3}} \right)$ = ${1 \over 6}$ $ \times $ ${1 \over 6}$ $ \times $ ${1 \over 2}$ = ${1 \over {72}}$

$\therefore$ ${E_1},$ ${E_2}$ and ${E_3}$ are not independent.

$\therefore$ Option (D) is correct.

$\bullet$ Probability of wining of T₁ against T₂ is = 1/2.

$\bullet$ Probability of drawing of T₁ against T₂ is = 1/6.

$\bullet$ Probability of losing of T₁ against T₂ is = 1/3.

$\bullet$3 points for win.

$\bullet$1 point for draw.

$\bullet$ 0 point for loss.

Here, P(X > Y) = P(T₁ win) P(T₁ win) + P(T₁ win) P(draw) + P(draw) P(T₁ win)

$ = \left( {{1 \over 2} \times {1 \over 2}} \right) + \left( {{1 \over 2} \times {1 \over 6}} \right) + \left( {{1 \over 6} \times {1 \over 2}} \right) = {5 \over {12}}$

$\bullet$ Probability of wining of T₁ against T₂ = 1/2.

$\bullet$ Probability of drawing of T₁ against T₂ is = 1/6.

$\bullet$ Probability of losing of T₁ against T₂ is = 1/3.

$\bullet$ 3 points for win.

$\bullet$ 1 point for draw.

$\bullet$ 0 point for loss.

P[X = Y] = P(draw) . P(draw) + P(T₁ win) P(T₂ win) + P(T₂ win) . P(T₁ win)

= (1/6 $\times$ 1/6) + (1/2 $\times$ 1/3) + (1/3 $\times$ 1/2) = 13/36

Let $P_1$ be the defective computers that are produced from plant $T_1$ and $P_2$ be that from plant $T_2$.

The total percentage of the defective computers produced is $7 \%$.

Now, $P_1=10 P$ and $P_2=P$.

The computers produced that are defective:

$ \begin{gathered} \frac{20}{100} \times P_1+\frac{80}{100} \times P_2=\frac{7}{100} \\\\ 20 P_1+80 P_2=7 \\\\ 200 P+80 P=7 \\\\ P=\frac{7}{280}=\frac{1}{40} \end{gathered} $

Now, the probability of the defective products is calculated as follows:

$ \frac{20}{100} P_1+\frac{80}{100} P_2=\frac{20}{100} \times \frac{1}{4}+\frac{80}{100} \times \frac{1}{40}=\frac{28}{400}=\frac{7}{100} $

The probability of producing NOT defective computers is

$ 1-\frac{7}{100}=\frac{93}{100} $

The probability that plant $T_2$ produces NOT defective computers is calculated as follows:

$ \frac{(80 / 100) \times(39 / 40)}{(93 / 100)}=\frac{78}{93} $

1^st ball can go any of the 3 boxes. So total choices for 1^st ball = 3

2^nd ball can also go any of the 3 boxes. So total choices for 2^nd ball = 3
.
.
.
.
12^th ball can go any of the 3 boxes. So total choices for 12^th ball = 3

Total choices for all 12 balls = $3 \times $$3 \times $$3 \times $.................12 times = 3¹².

Now question says choose 3 balls from 12 balls. So no of ways = ${}^{12}{C_3}$ ways.
And then put it in a box. No of ways we can put = ${}^{12}{C_3} \times 1$ ways.

Now we have 9 balls left and we have to put those 9 balls in the remaining 2 boxes.

Each ball can go to any of the 2 boxes, so for each ball there is 2 choices.

$\therefore$ Total ways for 9 balls = 2⁹

$\therefore$ Total ways we can put those 12 balls in the boxes = ${}^{12}{C_3} \times 1 \times {2^9}$

$\therefore$ Required probability = ${{{}^{12}{C_3} \times 1 \times {2^9}} \over {{3^{12}}}}$ = ${{55} \over 3}{\left( {{2 \over 3}} \right)^{11}}$

So option (C) is correct.

$\therefore$ P (drawing red ball from B₁) = ${1 \over 3}$

$ \Rightarrow \left( {{{{n_1} - 1} \over {{n_1} + {n_2} - 1}}} \right)\left( {{{{n_1}} \over {{n_1} + {n_2}}}} \right) + \left( {{{{n_2}} \over {{n_1} + {n_2}}}} \right)\left( {{{{n_1}} \over {{n_1} + {n_2} - 1}}} \right) = {1 \over 3}$

$ \Rightarrow {{n_1^2 + {n_1}{n_2} - {n_1}} \over {({n_1} + {n_2})({n_1} + {n_2} - 1)}} = {1 \over 3}$

Clearly, options (c) and (d) satisfy.

Box I : red balls $\to$ n₁

black balls $\to$ n₂

Box II : red balls $\to$ n₃

black balls $\to$ n₄

$P(R) = {1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}$

P(Box II / R) $ = {{{1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}} \over {{1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}}}$

$ = {1 \over {1 + \left( {{{{{{n_1}} \over {{n_1} + {n_2}}}} \over {{{{n_3}} \over {{n_3} + {n_4}}}}}} \right)}} = {1 \over 3}$

Thus, ${{{n_1}} \over {{n_1} + {n_2}}} = 2{{{n_3}} \over {{n_3} + {n_4}}}$

i.e. $2\left( {1 + {{{n_2}} \over {{n_1}}}} \right) = 1 + {{{n_4}} \over {{n_3}}}$ i.e. ${{{n_4}} \over {{n_3}}} - 2{{{n_2}} \over {{n_1}}} = 1$

$P(\overline {A \cup B} ) = {1 \over 6}$

or, $1 - P(A \cup B) = {1 \over 6}$

$\therefore$ $P(A \cup B) = 1 - {1 \over 6} = {5 \over 6};$ $P(A \cap B) = {1 \over 4}$; and $P(\overline A ) = {1 \over 4};$

$\therefore$ $P(A) = 1 - P(\overline A ) = 1 - {1 \over 4} = {3 \over 4}$

We know, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

or, ${5 \over 6} = {3 \over 4} + P(B) - {1 \over 4}$

or, $P(B) = {5 \over 6} - {1 \over 2} = {1 \over 3}$

Now, $P(A)\,.\,P(B) = {3 \over 4}.\,{1 \over 3} = {1 \over 4} = P(A \cap B)$

i.e., events A and B are mutually independent.

Since the probability of A and B are different, so they are not equally likely events.

Therefore, (A) is the correct option.

Given three Boxes:

Given, that a card is drow from each Box and $x_i$ denote the number on the card drawn from $i^{\text {tht }}$ box, $i=1,2,3$.

Sample space $={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105$.

Let A be the event when $x_1+x_2+x_3$ is odd.

Case I: When all $x_1, x_2, x_3$ are odd

No. of ways $={ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2.3 \cdot 4=24$

Case II: When $x_1$ is odd and $x_2, x_3$ are even

No. of ways $={ }^2 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=12$

Case III: When $x_2$ is odd and $x_1, x_3$ are even

No. of ways $={ }^1 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=9$

Case IV: When $x_3$ is odd and $x_1, x_2$ are even

No. of ways $={ }^1 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 . \mathrm{C}_1 \mathrm{C}_1=8$

No. of favorable cases for A is

$\begin{aligned} 24+12+9+8 & =53 \\ \text { Required Probability } & =\mathrm{P}(\mathrm{A})=\frac{53}{105} \end{aligned}$

Hint:

Recall that sum of three numbers is odd only when

(i) All three numbers are odd.

(ii) Two numbers are even and one is odd.

Given three Boxes :

Given, that a card is drow from each box and $x_i$ denote the number on the card drawn from $i^{\text {th }}$ box, $i=1,2,3$.

Sample Space $={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105$.

Let B be the event when $x_1, x_2, x_3$ are in A.P.

$\Rightarrow 2 x_2=x_1+x_3$

Here, $x_1+x_3$ is even number and for every even value of $x_1+x_3$ there is only one possible value of $x_2$

$x_1+x_2$ is even in two cases

Case I: When both $x_1$ and $x_2$ are even

No. of ways $={ }^1 \mathrm{C}_1$. ${ }^3 \mathrm{C}_{1=}=3$

Case II: When both $x_1$ and $x_2$ are odd

No. of ways $={ }^2 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2 \times 4=8$

Number of favorable cases for $\mathrm{B}=3+8=11$

Required probability $=\frac{11}{105}$

Hint:

$2 x_2=x_1+x_3$ indicate, $x_1+x_3$ is even and for every even value of $x_1+x_3$ there is only one possible value of $x_2$. Now, $x_1+x_3$ is even in two cases:

(i) When both $x_1$ and $x_2$ are even.

(ii) When both $x_1$ and $x_2$ are odd.

Given, 3 boys and 2 girls stand in a queue.

Sample space $n(s)=\left| \!{\underline {\, {5} \,}} \right.=120$

According to the given condition, following cases may arise.

BGGBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GGBBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GBGBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GBBGB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

BGBGB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

So, number of favorable ways $=5\left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.=60$

Required probability $=\frac{60}{120}=\frac{1}{2}$

Each question has 3 alternative and exactly one is correct.

$\therefore$ Probability of giving correct answer P(correct) = ${1 \over 3}$

$\therefore$ Probability of giving wrong answer P(wrong) = ${2 \over 3}$

Here student give 4 or more correct answer.

$\therefore$ Student give either 4 correct answer or 5 correct answer.

Using Binomial Distribution,

Required probability = ${}^5{C_4}{\left( {{1 \over 3}} \right)^4}{\left( {{2 \over 3}} \right)^1}$ + ${}^5{C_5}{\left( {{1 \over 3}} \right)^5}{\left( {{2 \over 3}} \right)^0}$ = ${{11} \over {{3^5}}}$

Let $E$ be the event of one white and one red balls $A_1, A_2, A_3$ be the events of both balls are from box $\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3$ respectively

Now, the probability of one white and one red both balls are drawn from box $B_2$ is $P\left(\frac{A_2}{E}\right)$.

$\Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2 \cap E\right)}{P(E)}$

$\begin{aligned} & \Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A_1}\right)+P\left(A_2\right).\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_2}\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_3}\right)} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}}{\frac{1}{3} \times \frac{{ }^1 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^3 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1}{{ }^{12} \mathrm{C}_2} } \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{6}}{\frac{1}{5}+\frac{1}{6}+\frac{2}{11}} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{55}{181} \\ \end{aligned}$

Hints:

$\Rightarrow$ Recall the Bay's theorem

$ \begin{gathered} P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A}\right)+P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)+P\left(A_3\right) \cdot P\left(\frac{E}{A_3}\right)}\\ \end{gathered}$

$\begin{aligned} & \text { Let } \mathrm{P} \text { be the probability of same colour balls } \\ & \text { drawn from each boxes } \mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3 \\ & \therefore \quad \mathrm{P}=\mathrm{P} \text { (all are white) }+\mathrm{P} \text { (all are red) }+\mathrm{P} \text { (all } \\ & \quad \text { are Black) } \\ & \Rightarrow P=\frac{1}{6} \times \frac{2}{9} \times \frac{3}{12}+\frac{3}{6} \times \frac{3}{9} \times \frac{4}{12}+\frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} \\ & \Rightarrow P=\frac{6}{648}+\frac{36}{648}+\frac{40}{648} \\ & \Rightarrow P=\frac{82}{648} \end{aligned}$

Hints:

The probability of one ball is drawn from each boxes $B_1, B_2, B_3$ is equal to the sum of probabilities of all are white balls, all are red balls and all are Black balls.

Let A, B, C, D are the events of solving a problem.

Given: $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{D})=\frac{1}{8}$

Also given A, B, C, D are independent events.

Let P is the probability of solved by at least one person.

$\Rightarrow \mathrm{P}=1-$ (Probability that none of the person can solve the problem)

$\begin{aligned} & \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}} \cap \overline{\mathrm{D}}) \\ & \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}}) \cdot \mathrm{P}(\overline{\mathrm{D}}) \\ & \Rightarrow \mathrm{P}=1-(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B}))(1-\mathrm{P}(\mathrm{C}))(1-\mathrm{P}(\mathrm{D})) \\ & \Rightarrow \mathrm{P}=1-\left(1-\frac{1}{2}\right)\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right) \\ & \Rightarrow \mathrm{P}=1-\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} \times \frac{7}{8} \\ & \Rightarrow \mathrm{P}=1-\frac{21}{256}\\ & \Rightarrow \mathrm{P}=\frac{235}{256} \end{aligned}$

Hints:

$\Rightarrow$ (i) If $A$ and $B$ are independent events, then

$\begin{aligned} & \rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \\ & \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}})=(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B})) \\ & \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\mathrm{B})=(1-\mathrm{P}(\mathrm{A})) \mathrm{P}(\mathrm{B}) \\ & \rightarrow \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot(1-\mathrm{P}(\mathrm{B})) \end{aligned}$

(ii) Probability of at least one person can solve the problem $= 1-$ Probability that none of the persons can solve that problem.

Given set S = $\left\{ {1,2,3,..8} \right\}$

Choosing 3 numbers from 8 numbers can be done ${{}^8{C_3}}$ ways.

Choosing 3 numbers from 8 numbers while minimum no is 3 can be done $1 \times {}^5{C_2}$ ways.

$\therefore$ Probablity P(min = 3) = ${{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,$

Choosing 3 numbers from 8 numbers while maximum no is 6 can be done $1 \times {}^5{C_2}$ ways.

$\therefore$ Probablity P(max = 6) = ${{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,$

Choosing 3 numbers from 8 numbers while minimum number 3 and maximum no is 6 can be done $1 \times {}^2{C_1} \times 1$ ways.

$\therefore$ $P\left( {\min = 3 \cap \max = 6} \right)$ = ${{1 \times {}^2{C_1} \times 1} \over {{}^8{C_3}}}$

The probability that their minimum is $3,$ given that their maximum is $6,$ is :
$P\left( {{{\min = 3} \over {\max = 6}}} \right)$

= ${{P\left( {\min = 3 \cap \max = 6} \right)} \over {P\left( {\max = 6} \right)}}$

= ${{{{{}^2{C_1}} \over {{}^8{C_3}}}} \over {{{{}^5{C_2}} \over {{}^8{C_3}}}}}$

= ${{{}^2{C_1}} \over {{}^5{C_2}}}$

= ${{1 \over 5}}$

For the given condition, the sample space $=6^4$

For favourable condition

Case I : For $D_4$ there are ${ }^6 C_1$ way. Now, it appears on any one of $D_1, D_2, D_3$ i.e. ${ }^3 C_1 \cdot 1$, for other two there are $5 \times 5$ ways.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1 \cdot 1 \cdot 5^2=450 \text { Ways }$

Case II : For $\mathrm{D}_4$ there are ${ }^6 \mathrm{C}_1$ ways. Now, it appears on any two of $D_1, D_2, D_3$ i.e. ${ }^3 \mathrm{C}_2 .1 .1$, for other one there are 5 ways.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_2 \cdot 1^2 \cdot 5=90 \text { Ways }$

Case III : For $\mathrm{D}_4$ there are ${ }^6 \mathrm{C}_1$ ways. Now, it appears on all i.e. $1^3$.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot 1^3=6$

Total number of favourable cases

$\begin{aligned} & =450+90+6 \\ & =546 \text { ways } \end{aligned}$

Thus, Probability $=\frac{546}{6^4}=\frac{91}{216}$

Let's analyze the given conditions and evaluate which options are correct.

Given:

1. $ P(X|Y) = \frac{1}{2} $


2. $ P(Y|X) = \frac{1}{3} $


3. $ P(X \cap Y) = \frac{1}{6} $

Now, let's proceed step by step through each option.

Option A: $ P(X \cup Y) = \frac{2}{3} $

We can use the formula for the union of two events:

$ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) $

To find $ P(X) $ and $ P(Y) $, we use the definitions of conditional probability:

$ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $

$ \frac{1}{2} = \frac{\frac{1}{6}}{P(Y)} $

$ P(Y) = \frac{1}{6} \times 2 = \frac{1}{3} $

Similarly,

$ P(Y|X) = \frac{P(X \cap Y)}{P(X)} $

$ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} $

$ P(X) = \frac{1}{6} \times 3 = \frac{1}{2} $

Now, substituting these values into the union formula:

$ P(X \cup Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} $

To add these fractions, we need a common denominator. The common denominator is 6:

$ P(X \cup Y) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} $

Hence, Option A is correct.

Option B: $ X $ and $ Y $ are independent

For two events $ X $ and $ Y $ to be independent, the following condition should hold:

$ P(X \cap Y) = P(X) \times P(Y) $

We already found that:

$ P(X) = \frac{1}{2} $

$ P(Y) = \frac{1}{3} $

$ P(X \cap Y) = \frac{1}{6} $

Checking the independence condition:

$ P(X) \times P(Y) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} $

Since this is equal to $ P(X \cap Y) $, $ X $ and $ Y $ are indeed independent.

Hence, Option B is correct.

Option C: $ X $ and $ Y $ are not independent

From the previous analysis, we have shown that $ X $ and $ Y $ are independent.

Hence, Option C is not correct.

Option D: $ P((X^c) \cap Y) = \frac{1}{3} $

To find $ P((X^c) \cap Y) $, we use the fact that:

$ P(Y) = P((X \cap Y) \cup (X^c \cap Y)) $

Since $ (X \cap Y) $ and $ (X^c \cap Y) $ are disjoint events, we can write:

$ P(Y) = P(X \cap Y) + P(X^c \cap Y) $

Given:

$ P(Y) = \frac{1}{3} $

$ P(X \cap Y) = \frac{1}{6} $

Substituting these,

$ \frac{1}{3} = \frac{1}{6} + P(X^c \cap Y) $

Solving for $ P(X^c \cap Y) $:

$ P(X^c \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} $

This contradicts Option D, as it says $ \frac{1}{3} $.

Hence, Option D is not correct.

In conclusion:

Option A and Option B are correct.

Option C and Option D are not correct.

$\begin{array}{r} \text { Given, } \mathrm{P}\left(\mathrm{X}_1\right)=\frac{1}{2} \Rightarrow \mathrm{P}\left(\overline{\mathrm{X}}_1\right)=1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{P}\left(\mathrm{X}_2\right)=\frac{1}{4} \Rightarrow \mathrm{P}\left(\bar{X}_2\right)=1-\frac{1}{4}=\frac{3}{4} \\ \mathrm{P}\left(\mathrm{X}_3\right)=\frac{1}{4} \Rightarrow \mathrm{P}\left(\bar{X}_3\right)=1-\frac{1}{4}=\frac{3}{4} \end{array}$

$\begin{aligned} & \text { Now, } P(X)=P\left(E_1 E_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)+P\left(E_1\right. \\ & \left.\overline{\mathrm{E}}_2 \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_1 \mathrm{E}_2 \overline{\mathrm{E}}_3\right) \\ & =\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} \\ \end{aligned}$

$\begin{aligned} & P(X)=\frac{1}{32}+\frac{1}{32}+\frac{3}{32}+\frac{3}{32} \\ & P(X)=\frac{1}{4} \end{aligned}$

(i) $\mathrm{P}\left(\frac{\mathrm{X}_1^c}{\mathrm{X}}\right)=\frac{\mathrm{P}\left(\mathrm{X}_1^c \cap \mathrm{X}\right)}{\mathrm{P}(\mathrm{X})}$

Now, $X_1^c \cap X$ indicates the event that the ship is operational but $\mathrm{E}_1$ is not functioning.

$\begin{aligned} & \therefore \quad \mathrm{P}\left(\mathrm{X}_1^c \cap \mathrm{X}\right)=\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{32} \\ & \therefore \mathrm{P}\left(\frac{\mathrm{X}_1^c}{\mathrm{X}}\right)=\frac{\frac{1}{32}}{\frac{1}{4}}=\frac{1}{8} \end{aligned}$

(ii) P (Exactly two engines of the ship are functioning $\mid X$)

$\begin{aligned} & =\frac{\mathrm{P}\left(\overline{\mathrm{E}}_1 \mathrm{E}_2 \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_1 \overline{\mathrm{E}}_2 \mathrm{E}_3\right)+1\left(\mathrm{E}_1 \mathrm{E}_2 \overline{\mathrm{E}}_3\right)}{\mathrm{P}(\mathrm{X})} \\ & =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}} \\ & =\frac{7}{8} \end{aligned}$

$\begin{aligned} \text { (iii) } & P\left(X \mid X_2\right)=\frac{P\left(X \cap X_2\right)}{P\left(X_2\right)} \\ & =\frac{P\left(E_1 E_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)+P\left(E_1 E_2 \bar{E}_3\right)}{P\left(X_2\right)} \\ & =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}} \\ & =\frac{5}{8} \end{aligned}$

$\begin{aligned} \text { (iv) } & P\left(X \mid X_1\right)=\frac{P\left(X \cap X_1\right)}{P\left(X_1\right)} \\ = & \frac{P\left(E_1 E_2 E_3\right)+P\left(E_1 \bar{E}_2 E_3\right)+P\left(E_1 E_2 \bar{E}_3\right)}{P\left(X_1\right)} \\ = & \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}} \\ = & \frac{7}{16} \end{aligned}$

Here is 5 trials.
So according to Bernoulli trial n = 5

P( at least one failure) = 1 - P( no failure)

According to question,
1 - P( no failure) $ \ge {{31} \over {32}}$

$ \Rightarrow $ 1 - P( x = 5 ) $ \ge {{31} \over {32}}$

[Note: no failure = all success]

$ \Rightarrow $ 1 - ${}^5{C_5}\,{p^5}$ $ \ge {{31} \over {32}}$

$ \Rightarrow $ 1 - ${p^5}$ $ \ge {{31} \over {32}}$

$ \Rightarrow $ ${p^5}$ $ \le $ ${1 \over {32}}$

$ \Rightarrow $ $p$ $ \le $ ${1 \over {2}}$

$\therefore$ $p$ $ \in $ $\left[ {0,{1 \over 2}} \right]$

Note: $p$ can not be less than 0 as probability is always $\ge$ 0 and $\le$ 1.

Given that $C \subset D$ means $C$ is present entirely inside $D$. Which is shown below.

$P\left( {{C \over D}} \right)$ = ${{P\left( {C \cap D} \right)} \over {P\left( D \right)}}$ = ${{P\left( C \right)} \over {P\left( D \right)}}$

As $C \cap D$ means common part of events C and D which is equal to C.

$0 \le P\left( D \right) \le 1$

$\therefore$ ${{P\left( C \right)} \over {P\left( D \right)}} \ge P\left( C \right)$

Note: Here we are dividing with ${P\left( D \right)}$ which is $ \le 1$ and $ \ge 0$, as we know on dividing with a number n in the range $0 \le n \le 1$ we get always more than or equal to the original number.

$P\left( {{H \over W}} \right) = {{P(W/H) \times P(H)} \over {P(W/T)\,.\,P(T) + (W/H)\,.\,P(H)}}$

$ = {{{1 \over 2}\left( {{3 \over 5} \times 1 + {2 \over 5} \times {1 \over 2}} \right)} \over {23/30}} = {{12} \over {23}}$

H $\to$ One ball from U₁ to U₂.

T $\to$ Two balls from U₁ to U₂.

E : One ball drawn from U₂.

P/W from

${U_2} = {1 \over 2} \times \left( {{3 \over 5} \times 1} \right) + {1 \over 2} \times \left( {{2 \over 5} \times {1 \over 2}} \right) + {1 \over 2} \times \left( {{{{}^3{C_2}} \over {{}^5{C_2}}} \times 1} \right) + {1 \over 2} \times \left( {{{{}^2{C_2}} \over {{}^5{C_2}}} \times {1 \over 3}} \right) + {1 \over 2} \times \left( {{{{}^3{C_1}\,.\,{}^2{C_1}} \over {{}^5{C_2}}} \times {2 \over 3}} \right) = {{23} \over {30}}$

Let $P(E) = e$ and $P(F) = f$.

$P(E \cup F) - P(E \cap F) = {{11} \over {25}}$

$ \Rightarrow e + f - 2ef = {{11} \over {25}}$ ...... (1)

$P(\overline E \cap \overline F ) = {2 \over {25}}$

$ \Rightarrow (1 - e)(1 - f) = {2 \over {25}}$

$ \Rightarrow 1 - e - f + ef = {2 \over {25}}$ ...... (2)

From Eqs. (1) and (2), we get

$ef = {{12} \over {25}}$ and $e + f = {7 \over 5}$

Solving, we get

$e = {4 \over 5},\,f = {3 \over 5}$ or $e = {3 \over 5},\,f = {4 \over 5}$

Out of nine balls three balls can be chosen = ${}^9{C_3}$ ways

$\therefore$ Sample space = ${}^9{C_3}$ = ${{9!} \over {3!6!}}$ = ${{9 \times 8 \times 7} \over 6}$ = 84

According to the question, all three ball should be different. So out of 3 red balls 1 is chosen and out of 4 blue 1 is chosen and out of 2 green 1 is chosen.

$\therefore$ Total cases = ${}^3{C_1} \times {}^4{C_1} \times {}^2{C_1}$ = 3 $ \times $ 4 $ \times $ 2 = 24

$\therefore$ Probability = ${{24} \over {84}}$ = ${{2} \over {7}}$

Four numbers can be chosen ${}^{20}{C_4}$ ways.

When common difference d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets
So when d = 1 then 17 different AP's are possible with 4 numbers.

Now let's create a table of all possible sets -

Common Difference (d)	Possible Sets	No of AP
d = 1	(1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20)	17
d = 2	(1, 3, 5, 7) (2, 4, 6, 8) (3, 5, 7, 9) ................. (14, 16, 18, 20)	14
d = 3	(1, 4, 7, 10) (2, 5, 8, 11) (3, 6, 9, 12) ................. (11, 14, 17, 20)	11
d = 4	(1, 5, 9, 13) (2, 6, 10, 14) (3, 7, 11, 15) ................. (8, 12, 16, 20)	8
d = 5	(1, 6, 11, 16) (2, 7, 12, 17) (3, 8, 13, 18) (4, 9, 14, 19) (5, 10, 15, 20)	5
d = 6	(1, 7, 13, 19) (2, 8, 14, 20)	2

$\therefore$ Total no of AP = 17 + 14 + 11 + 8 + 5 + 2 = 57

$\therefore$ Required probability = ${{57} \over {{}^{20}{C_4}}}$ = ${1 \over {85}}$

$\therefore$ Statement - 1: is true.

$\therefore$ Statement - 2: is false as common difference can also be $ \pm 6$.

Sample space A dice is thrown thrice, $n(s) = 6 \times 6 \times 6$.

Favorable events ${\omega ^{{r_1}}} + {\omega ^{{r_2}}} + {\omega ^{{r_3}}} = 0$

i.e. $({r_1},{r_2},{r_3})$ are ordered 3-triples which can take values,

$\left. {\matrix{ {(1,2,3),} & {(1,5,3),} & {(4,2,3),} & {(4,5,3)} \cr {(1,2,6),} & {(1,5,6),} & {(4,2,6),} & {(4,5,6)} \cr } } \right\}$

i.e. 8 ordered pairs and each can be arranged in 3! ways = 6

$\therefore$ $n(E) = 8 \times 6$

$ \Rightarrow P(E) = {{8 \times 6} \over {6 \times 6 \times 6}} = {2 \over 9}$

From the tree-diagram it follows that

$P({B_G}) = {{46} \over {80}}$

$P({B_G}|G) = {{10} \over {16}} = {5 \over 8}$

$\therefore$ $P({B_G} \cap G) = {5 \over 8} \times {4 \over 5} = {1 \over 2}$

$P(G|{B_G}) = {{{1 \over 2}} \over {P({B_G})}} = {1 \over 2} \times {{80} \over {46}} = {{20} \over {23}}$

Sample space = {00, 01, 02, 03, ..........49} = 50 tickets

n(S) = 50

n(Sum = 8) = { 08, 17, 26, 35, 44 } = 5

n(Product = 0) = { 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40 } = 14

$\therefore$ Probability when product is 0 = P(Product = 0) = ${14 \over {50}}$

n(Sum = 8 $ \cap $ Product = 0) = { 08 } = 1

$\therefore$ Probability when sum is 8 and product is 0 = P(Sum = 8 $ \cap $ Product = 0) = ${1 \over {50}}$

Required probability,
$P\left( {{{Sum = 8} \over {Product = 0}}} \right)$

= ${{P\left( {Sum = 8 \cap Product = 0} \right)} \over {P\left( {Product = 0} \right)}}$

= ${{{1 \over {50}}} \over {{{14} \over {50}}}}$

=${1 \over {14}}$

$\therefore$ Option (D) is correct.

Given that, no of trials = n
Probability of success (p) = ${{1 \over 4}}$
$\therefore$ Probability of no success = 1 - ${{1 \over 4}}$ = ${{3 \over 4}}$

As we know, probability of at least one success = 1 - probability of no success

$\therefore$ According to the question,

1 - (Probability of no success in n trials) $\ge$ ${9 \over {10}}$

$ \Rightarrow $ 1 - P(x = 0) $\ge$ ${9 \over {10}}$

$ \Rightarrow $ 1 - ${}^n{C_0}$${\left( {{1 \over 4}} \right)^0}$${\left( {{3 \over 4}} \right)^n}$ $\ge$ ${9 \over {10}}$

$ \Rightarrow $ 1 - ${\left( {{3 \over 4}} \right)^n}$ $\ge$ ${9 \over {10}}$

$ \Rightarrow $ ${\left( {{3 \over 4}} \right)^n}$ $\le$ ${1 \over {10}}$

$ \Rightarrow $ ${\left( {{4 \over 3}} \right)^n}$ $\ge$ $10$

[Taking log on both sides]

$ \Rightarrow $ $n\left( {\log _{10}^4 - \log _{10}^3} \right)$ $\ge$ ${\log _{10}^{10}}$

$ \Rightarrow $ $n\left( {\log _{10}^4 - \log _{10}^3} \right)$ $\ge$ $1$

$ \Rightarrow $ $n$ $\ge$ ${1 \over {\left( {\log _{10}^4 - \log _{10}^3} \right)}}$

$\therefore$ Option (D) is correct.

The required probability is $P(X = 3)$

$\left( {{5 \over 6}} \right)\left( {{5 \over 6}} \right){1 \over 6} = {{25} \over {216}}$

The probability $P(X \ge 3)$ is nothing but the probability of $P(X \le 2)$:

${1 \over 6} + {5 \over 6} \times {1 \over 6} = {{11} \over {36}}$

The required probability is

$1 - {{11} \over {36}} = {{25} \over {36}}$

$X \ge 6$: The probability for

${{{5^5}} \over {{6^6}}} + {{{5^6}} \over {{6^7}}}\, + \,...\, + \,\infty = {{{5^5}} \over {{6^6}}}\left( {{1 \over {1 - 5/6}}} \right) = {\left( {{5 \over 6}} \right)^5}$

For $X>3$, we have

${{{5^3}} \over {{6^4}}} + {{{5^4}} \over {{6^5}}}\, + {{{5^5}} \over {{6^6}}}\, + ...\, + \,\infty = {\left( {{5 \over 6}} \right)^3}$

Hence, the conditional probability is

${{{{(5/6)}^6}} \over {{{(5/6)}^3}}} = {{25} \over {36}}$

No of outcome for a die = { 1, 2, 3, 4, 5, 6 }
According to the question,

A = { 4, 5, 6 }

$\therefore$ P(A) = ${3 \over 6}$

B = { 1, 2, 3, 4 }

$\therefore$ P(A) = ${4 \over 6}$

A $ \cap $ B = { 4 }

So P(A $ \cap $ B) = ${1 \over 6}$

We know, $P\left( {A \cup B} \right)$ = $P\left( A \right)$ + $P\left( B \right)$ - $P\left( {A \cap B} \right)$

$\therefore$ $P\left( {A \cup B} \right)$ = ${3 \over 6}$ + ${4 \over 6}$ - ${1 \over 6}$ = ${{7 - 1} \over 6}$ = ${6 \over 6}$ = 1

$\therefore$ Option (C) is correct.

Given that,

$P\left( {{A \over B}} \right) = {1 \over 2}$

$ \Rightarrow $ ${{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$ = ${1 \over 2}$.............. equation (1)

$P\left( {{B \over A}} \right) = {2 \over 3}$

$ \Rightarrow $ ${{P\left( {A \cap B} \right)} \over {P\left( A \right)}}$ = ${2 \over 3}$.............. equation (2)

Dividing equation (1) by equation (2) we get,

${{P\left( A \right)} \over {P\left( B \right)}}$ = ${3 \over 4}$

$ \Rightarrow $ ${P\left( B \right)}$ = ${4 \over 3}$ $ \times $ ${P\left( A \right)}$
                   = ${4 \over 3}$ $ \times $ ${1 \over 4}$
                   = ${1 \over 3}$

$\therefore$ Option (B) is correct.