Probability

$1 - {}^n{C_0}{\left( {{1 \over 3}} \right)^0}{\left( {{2 \over 3}} \right)^n} > {5 \over 6}$

${1 \over 6} > {\left( {{2 \over 3}} \right)^n}\,\, \Rightarrow \,\,0.1666 > {\left( {{2 \over 3}} \right)^n}$

${n_{\min }} = 5$

$P\left( A \right) = {1 \over 2} \times {{11} \over {36}} + {1 \over 2} \times {2 \over 9} = {{19} \over {72}}$

5 Red and 2 green balls

P(one red ball) = ${5 \over 7}$

P(one green ball) = ${2 \over 7}$

Case I :

If drawn ball is green than a red ball is added

$\left( {\matrix{ {6{\mathop{\rm Re}\nolimits} d} \cr {1\,Green} \cr } } \right)$ P (red ball) = ${6 \over 7}$

Case II :

If drawn ball is red than a green ball is added

$\left( {\matrix{ {4{\mathop{\rm Re}\nolimits} d} \cr {3\,Green} \cr } } \right)$ P (red ball) = ${4 \over 7}$

P (2^nd red ball) = ${5 \over 7}$ $ \times {4 \over 7} + {2 \over 7} \times {6 \over 7}$ = ${{32} \over {49}}$

P (X = 1) means out of two drawn cards one card is ace.

and P(X = 2) means both the drawn cards are ace.

$ \therefore $  P(X = 1) = first card is ace or 2nd card is ace.

= A _ $+$ _ A

= ${4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}$

= $2 \times {4 \over {52}} \times {{48} \over {52}}$

P(X = 2) = First and second both cards arc ace.

= A A

= ${4 \over {52}} \times {4 \over {52}}$

$ \therefore $  P(X = 1) + P(X = 2)

= $2 \times {4 \over {52}} \times {{48} \over {52}} + {4 \over {52}} \times {4 \over {52}}$

= ${{25} \over {169}}$

Here, $P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}$

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

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

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

= ${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}$ ($ \because $$\left. {P\left( {A \cap B \cap C} \right) = 0} \right)$

= ${{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}$

[$ \because $ A, B and C are independent events]

= 1 - P(A) - P(B)

= $P\left( {\overline A } \right)$ - P(B) or $P\left( {\overline B } \right)$ - P(A)

Let the number of children in each family be x.

Thus the total number of children in both the families are 2x

Now, it is given that 3 tickets are distributed amongst the children of these two families.

Thus, the probability that all the three tickets go to the children in family B

= ${{{}^x{C_3}} \over {{}^{2x}{C_3}}}$ = ${1 \over {12}}$

$ \Rightarrow $ $\,\,\,$ ${{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}$ = ${1 \over {12}}$

$ \Rightarrow $  ${{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}$ = ${1 \over 6}$

Thus, the number of children in each family is 5.

If we follow path 1, then probability of getting 1st ball black $ = {6 \over {10}}$ and probability of getting 2nd ball red when there is 4 R and 8 B balls = ${4 \over {12}}$.

So, the probability of getting 1st ball black and 2nd ball red = ${6 \over {10}} \times {4 \over {12}}$.

If we follow path 2, then the probability of getting 1st ball red $ = {4 \over {10}}$ and probability of getting 2nd ball red when in the bag there is 6 red and 6 black balls = ${6 \over {12}}$

$\therefore\,\,\,$ Probability of getting 2nd ball as red

$ = {6 \over {10}} \times {4 \over {12}} + {4 \over {10}} \times {6 \over {12}}$

$ = {1 \over 5} + {1 \over 5}$

$ = {2 \over 5}$

P(X getting head) = p

$ \therefore $ P(X getting tail) = 1 - p

P(Y getting head) = P(Y getting tail) = ${1 \over 2}$

P(X wins) = p + (1 - p)${1 \over 2}$p + (1 - p)${1 \over 2}$(1 - p)${1 \over 2}$p + ...

= ${p \over {1 - \left( {{{1 - p} \over 2}} \right)}}$

= ${{2p} \over {1 + p}}$

P(Y win) = (1 - p)${1 \over 2}$ + (1 - p)${1 \over 2}$(1 - p)${1 \over 2}$ + ...

= $\left( {{{1 - p} \over 2}} \right).{p \over {1 - \left( {{{1 - p} \over 2}} \right)}} = {{1 - p} \over {1 + p}}$

According to question,

P(X wins) = P(Y wins)

$ \therefore $ ${{2p} \over {1 + p}}$ = ${{1 - p} \over {1 + p}}$

$ \Rightarrow $ 3p = 1

$ \Rightarrow $ p = ${1 \over 3}$

Probability of drawing a white ball and then a red ball

from bag B is given by ${{{}^4{C_1} \times {}^2{C_1}} \over {{}^9{C_2}}}$ = ${2 \over 9}$

Probability of drawing a white ball and then a red ball

from bag A is given by ${{{}^2{C_1} \times {}^3{C_1}} \over {{}^7{C_2}}}$ = ${2 \over 7}$

Hence, the probability of drawing a white ball and then

a red ball from bag B = ${{{2 \over 9}} \over {{2 \over 7} + {2 \over 9}}}$ = ${{2 \times 7} \over {18 + 14}}$ = ${7 \over {16}}$

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)

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

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.

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.

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

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.

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

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

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

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

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

= ${{1 \over 5}}$

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.

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 = {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.

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.

Probability of getting score 9 in a single throw

$ = {4 \over {36}} = {1 \over 9}.$

Probability of getting score 9 exactly twice

$ = {}^3{C_2} \times {\left( {{1 \over 9}} \right)^2} \times {8 \over 9} = {8 \over {243}}.$

The desired probability

= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + .......

= 0.14 [1 + (0.56) + (0.56)² + .......]

= 0.14 $\left( {{1 \over {1 - 0.56}}} \right) = {{0.14} \over {0.44}} = {7 \over {22}}$ = 0.32

Required probability

$ = P(X = 0) + P(X = 1)$

$ = {{{e^{ - 5}}} \over {0!}}{5^0} + {{{e^{ - 5}}} \over {1!}}{5^1}$

$ = {e^{ - 5}} + 5{e^{ - 5}}$

$ = {6 \over {{e^5}}}$

Person 1st has three options to apply.

Similarly, person 2nd has three options t apply

and person 3rd has three options to apply.

Total cases = 3³

Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3)

So, probability $ = {3 \over {{3^3}}} = {1 \over 9}$.

Given that,

$P(\overline {A \cup B} ) = {1 \over 6}$, $P(A \cap B) = {1 \over 4}$, $P(\overline A ) = {1 \over 4}$

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

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

$ \Rightarrow 1 - P(A) - P(B) + P(A \cap B) = {1 \over 6}$

$ \Rightarrow P(\overline A ) - P(B) + {1 \over 4} = {1 \over 6}$

$ \Rightarrow P(B) = {1 \over 4} + {1 \over 4} - {1 \over 6}$

$ \Rightarrow P(B) = {1 \over 3}$ and $P(A) = {3 \over 4}$

Clearly, $P(A \cap B) = P(A)P(B)$,

so, the events A and B are independent events but not equally likely.

In a position distribution,

$P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}$ ($\lambda$ = mean).

Now, $P(X = r > 1.5) = P(2) + P(3) + $ ..... $\infty$

$ = 1 - \{ P(0) + P(1)\} $

$ = 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}}}$

The probability of speaking truth by A, P(A) = ${4 \over 5}$. The probability of not speaking truth by A, P($\overline A $) = 1 $-$ ${4 \over 5} = {1 \over 5}$.

The probability of speaking truth by B, P(B) = ${3 \over 4}$. The probability of not speaking truth of B, P($\overline B $) $ = {1 \over 4}$.

The probability that they contradict each other

$ = P(A) \times P(\overline B ) + P(\overline A ) \times P(B)$

$ = {4 \over 5} \times {1 \over 4} + {1 \over 5} \times {3 \over 4}$

$ = {1 \over 5} + {3 \over {20}} = {7 \over {20}}$

Given that mean = 4 = np = 4 and variance = 2

npq = 2 $\Rightarrow$ 4q = 2

$ \Rightarrow q = {1 \over 2}$

$\therefore$ $p = 1 - q = 1 - {1 \over 2} = {1 \over 2}$

Also, n = 8

Probability of 2 successes $ = P(X = 2) = {}^8{C_2}{p^2}{q^6}$

$ = {{8!} \over {2! \times 6!}} \times {\left( {{1 \over 2}} \right)^2} \times {\left( {{1 \over 2}} \right)^6} = 28 \times {1 \over {{2^8}}}$

$ = {{28} \over {256}}$

X : Mr. A selected wining horse

$\overline X $ : Mr. A did not select wining horse

Mr. A selected two horses, now probability of not wining the first horse which Mr. A choses = ${4 \over 5}$

And probability of not wining the second horse also which Mr. A choses = ${3 \over 4}$ (Here Mr. A out of remaining 4 horses choses one horse among 3 horses which did not win)

$\therefore$ $P(\overline X ) = {4 \over 5} \times {3 \over 4}$

$\therefore$ $P(X) = 1 - P(\overline X )$

$ = 1 - {4 \over 5} \times {3 \over 4}$

$ = 1 - {3 \over 5}$

$ = {2 \over 5}$

Mean = $np$ = 4
Variance = $npq$ = 2

$ \Rightarrow 4 \times q$ = 2
$ \Rightarrow$ $q$ = ${1 \over 2}$
p = 1 - q = 1 - ${1 \over 2}$ = ${1 \over 2}$
n = 8

Now P(X = 1) = ${}^8{C_1}{\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{8 - 1}}$
                        = $8 \times {1 \over {{2^8}}}$
                        = ${1 \over {32}}$

Given $P\left( A \right) = {{3x + 1} \over 3},$ $P\left( B \right) = {{1 - x} \over 4}$ and $P\left( C \right) = {{1 - 2x} \over 2}$

We know for any event X, $0 \le P\left( X \right) \le 1$

$\therefore$ $0 \le {{3x + 1} \over 3} \le 1$
$ \Rightarrow - 1 \le 3x \le 2$
$ \Rightarrow - {1 \over 3} \le x \le {2 \over 3}$

$0 \le {{1 - x} \over 4} \le 1$
$ \Rightarrow - 3 \le x \le 1$

$0 \le {{1 - 2x} \over 2} \le 1$
$ \Rightarrow - 1 \le 2x \le 1$
$ \Rightarrow - {1 \over 2} \le x \le {1 \over 2}$

In the question given that A, B and C are mutually exclusive
So $P\left( {A \cup B \cup C} \right)$ = $P\left( {A} \right)$ + $P\left( {B} \right)$ + $P\left( {C} \right)$
 
$ \Rightarrow P\left( {A \cup B \cup C} \right)$ = ${{3x + 1} \over 3}$ + ${{1 - x} \over 4}$ + ${{1 - 2x} \over 2}$

$\therefore$ 0 $ \le $ ${{3x + 1} \over 3}$ + ${{1 - x} \over 4}$ + ${{1 - 2x} \over 2}$ $ \le $ 1

0 $ \le $ 13 - 3$x$ $ \le $ 12

$ \Rightarrow {1 \over 3} \le x \le {{13} \over 3}$

From all those relations, we get

$\max \left\{ { - {1 \over 3}, - 3, - {1 \over 2},{1 \over 3}} \right\}$ $ \le $ $x$ $ \le $ $\min \left\{ {{2 \over 3},1,{1 \over 2},{{13} \over 3}} \right\}$

So, ${1 \over 3} \le x \le {1 \over 2}$

$ \Rightarrow $ $ \Rightarrow x \in \left[ {{1 \over 3},{1 \over 2}} \right]$

Given $P\left( A \right) = {1 \over 2}$, $P\left( B \right) = {1 \over 3}$, $P\left( C \right) = {1 \over 4}$

So, $P\left( {\overline A } \right) = {1 \over 2}$ (Probablity that the problem can't be solve by A)
$P\left( {\overline B } \right) = {2 \over 3}$ (Probablity that the problem can't be solve by B)
and $P\left( {\overline C } \right) = {3 \over 4}$ (Probablity that the problem can't be solve by C)

Now the probablity that the problem is solved by any one student of A, B and C = 1 - the probablity that the problem is solved by none of the students of A, B and C

$P\left( {A \cup B \cup C} \right)$ = 1 - $P\left( {\overline A } \right)P\left( {\overline B } \right)P\left( {\overline C } \right)$

$ = 1 - {1 \over 2} \times {2 \over 3} \times {3 \over 4}$
$=$ ${3 \over 4}$

Total no of trials = 5
$\therefore$ n = 5
Odd no possibe are = {1, 3, 5} =3
Sample space = {1, 2, 3, 4, 5, 6} = 6

Probablity of getting odd number = ${3 \over 6}$ = ${1 \over 2}$
$\therefore$ p = ${1 \over 2}$

Formula for variance = npq
where q = 1 - p = 1 - ${1 \over 2}$ = ${1 \over 2}$
$\therefore$ Variance = $5 \times {1 \over 2} \times {1 \over 2}$ = ${5 \over 4}$

Given $P\left( {A \cup B} \right) = 3/4$,
$P\left( {A \cap B} \right) = 1/4,$
$P\left( {\overline A } \right) = 2/3$

We know, $P\left( A \right)$ = 1 - $P\left( {\overline A } \right)$
$\therefore$ $P\left( A \right)$ = 1 - ${2 \over 3}$ = ${1 \over 3}$

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

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

$ \Rightarrow $ 1 = ${1 \over 3}$ + $P\left( B \right)$

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

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

So $P\left( {\overline A \cap B} \right)$ = ${2 \over 3}$ - ${1 \over4}$ = ${5 \over 12}$