Probability

We have, $P(A \cup B)=\frac{3}{4}$,

$ P(A \cap B)=\frac{1}{4} \text { and } P(\bar{A})=\frac{2}{3} $

We know that,

$ \begin{aligned} & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow P(B)-P(A \cap B)=P(A \cup B)-P(A) \\ & \Rightarrow P(\bar{A} \cap B)=P(A \cup B)-P(A) \\ & \quad=\frac{3}{4}-\left(1-\frac{2}{3}\right)=\frac{3}{4}-\frac{1}{3}=\frac{5}{12} \end{aligned} $

We have, $P(X)=\frac{1}{3}, P(X / Y)=\frac{1}{2}$

and $P(Y / X)=\frac{2}{5}$

Now, $\quad P(Y / X)=\frac{2}{5}$

$ \begin{aligned} & \Rightarrow \quad \frac{P(Y \cap X)}{P(X)}=\frac{2}{5} \\ & \Rightarrow P(Y \cap X)=\frac{2}{5} \times \frac{1}{3}=\frac{2}{15} \\ & \text { Again, } P(X / Y)=\frac{P(X \cap Y)}{P(Y)} \\ & \qquad \frac{1}{2}=\frac{\frac{2}{15}}{P(Y)} \Rightarrow P(Y)=\frac{4}{15} \end{aligned} $

$ \begin{aligned} & \text { Also, } P(X \cup Y)=P(X)+P(Y)-P(X \cap Y) \\ & \quad=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{5+4-2}{15}=\frac{7}{15} \\ & \therefore P(\bar{X} / Y)=\frac{P(\bar{X} \cap Y)}{P(Y)} \\ & \quad=\frac{P(Y)-P(X \cap Y)}{P(Y)} \\ & \quad=1-\frac{P(X \cap Y)}{P(Y)}=1-\frac{\frac{2}{15}}{\frac{4}{15}}=1-\frac{1}{2} \\ & =\frac{1}{2} \end{aligned} $

We have,

$ P(A)=\frac{4}{9}, P(A \cap \bar{B})=\frac{3}{7} $

Now, $P(A \cap \bar{B})=P(A)-P(A \cap B)$

$ \begin{aligned} & \Rightarrow \quad P(A \cap B)=P(A)-P(A \cap \bar{B}) \\ & \quad=\frac{4}{9}-\frac{3}{7}=\frac{28-27}{63}=\frac{1}{63} \\ & \therefore P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{\left(\frac{1}{63}\right)}{\left(\frac{4}{9}\right)}=\frac{1}{28} \end{aligned} $

We have,

$ n=4, p=\frac{20}{100}=\frac{1}{5}, q=1-p=1-\frac{1}{5}=\frac{4}{5} $

∴ Required probability $=P(x<2)$

$ \begin{aligned} & =P(x=0)+P(x=1) \\ & ={ }^4 C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^4+{ }^4 C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^3 \\ & =\left(\frac{4}{5}\right)^4+\left(\frac{4}{5}\right)^4 \\ & =2\left(\frac{4}{5}\right)^4=0.8192 \end{aligned} $

$ \begin{aligned} & \text { We have, } \lambda=\frac{60}{600}=0.1 \\ & \begin{aligned} \therefore \quad P(X & =x)=\frac{e^{-\lambda}(\lambda)^x}{x!} \\ & =\frac{e^{-0.1}(0.1)^x}{x!} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Required probability }\\ &\begin{aligned} & =P(x=0)+P(x=1)+P(x=2) \\ & =e^{-0.1}\left[\frac{(0.1)^0}{0!}+\frac{(0.1)^1}{1!}+\frac{(0.1)^2}{2!}\right] \\ & =e^{-0.1}\left[1+\frac{1}{10}+\frac{1}{200}\right] \\ & =e^{-0.1}\left[\frac{200+20+1}{200}\right] \\ & =e^{-0.1}\left(\frac{221}{200}\right)=\frac{1}{e^{0.1}}\left(\frac{221}{200}\right) \end{aligned} \end{aligned} $

When two dice are thrown then sample space will {(1, 1), (2, 2) ....... (6, 6)} contain total 36 elements number of cases.

Then the expectation will be ${6 \over {36}} \times 15 \times {4 \over {36}} \times 12 - {{26} \over {36}} \times 6$

${{90 + 48 - 156} \over {36}} = - {1 \over 2}$ = $ {1 \over 2}$ loss

There are 50 questions in an exam.

So Probability of each question to be correct is p = ${4 \over 5}$

and also probability of each question to be incorrect is q = ${1 \over 5}$

Let Y is the number of correct question in 50 questions, hence required probability is

${\left( {{4 \over 5}} \right)^{50}} + {}^{50}{C_1}\left( {{1 \over 5}} \right){\left( {{4 \over 5}} \right)^{49}}$

= ${\left( {{4 \over 5}} \right)^{49}} + \left( {{4 \over 5} + 10} \right)\left( {{4 \over 5}} \right)$

= ${{54} \over 5}{\left( {{4 \over 5}} \right)^{49}}$

Choosing vertices of a regular hexagon alternate, here A₁, A₃, A₅ or A₂, A₄, A₆ will result in an equilateral triangle.

Hence the required probability = ${2 \over {{}^6{C_3}}}$ = ${1 \over {10}}$

Let number of trials be n and probability of success = p, probability of failure = q

Given np = 8, npq = 4

$ \Rightarrow $ q = ${1 \over 2}$, p = ${1 \over 2}$, n = 16 (as p + q = 1)

p$\left( {x \le 2} \right)$ = ${{{}^{16}{C_0} + {}^{16}{C_1} + {}^{16}{C_2}} \over {{2^{16}}}} = {{1 + 16 + 120} \over {{2^{16}}}} = {{137} \over {{2^{16}}}}$

$1 - {\left( {{1 \over 2}} \right)^n} > {{99} \over {100}}$

${\left( {{1 \over 2}} \right)^n} < {1 \over {100}}$

$ \Rightarrow $ n = 7

A = At least two girls

B = All girls

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

$ \Rightarrow {{P(B)} \over {P(A)}} = {{{{\left( {{1 \over 4}} \right)}^2}} \over {1 - {}^4{C_0}{{\left( {{1 \over 2}} \right)}^4} - {}^4{C_1}{{\left( {{1 \over 2}} \right)}^4}}}$

$ \Rightarrow {1 \over {16 - 1 - 4}} = {1 \over {11}}$

Let four persons are A, B, C and D.

Probablity of hitting a target by them,

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

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

P(C) = ${1 \over 4}$

P(D) = ${1 \over 8}$

Probablity of hitting target atleast once = 1 - Probablity of not hitting by anybody

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

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

= 1 - ${1 \over 2}.{2 \over 3}.{3 \over 4}.{7 \over 8}$

= ${{25} \over {32}}$

Probablity of getting head P(H) = ${1 \over 2}$

Probablity of getting tail P(T) = 1 - ${1 \over 2}$ = ${1 \over 2}$

Probability of observing at least one head out of n tosses

= 1 - Probability of observing no head occurs out of n tosses

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

According to the question,

1 - ${\left( {{1 \over 2}} \right)^n}$ $ \ge $ ${{90} \over {100}}$

$ \Rightarrow $ ${\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$

$ \Rightarrow $ ${2^n} \ge 10$

As 2³ = 8

2⁴ = 16

$ \therefore $ Minimum value of n = 4

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

As A $ \subset $ B,

then P(A$ \cap $B) = P(A)

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

As P(B) $ \le $ 1

$ \therefore $ ${{P\left( A \right)} \over {P\left( B \right)}}$ $ \ge $ P(A)

A $ \to $ opted NCC

B $ \to $ opted NSS

$ \therefore $  P (nither A nor B) $=$ ${{10} \over {60}} = z{1 \over 6}$

Expected Gain/Loss =

= w $ \times $ 100 + Lw($-$ 50 + 100)

       + L²w ($-$ 50 $-$ 50 + 100) + L³($-$ 150)

= ${1 \over 3} \times 100 + {2 \over 3}.{1 \over 3}\left( {50} \right) + {\left( {{2 \over 3}} \right)^2}\left( {{1 \over 3}} \right)\left( 0 \right)$

        $ + {\left( {{2 \over 3}} \right)^3}\left( { - 150} \right) = 0$

here w denotes probability that outcome 5 or 6 (w = ${2 \over 6} = {1 \over 3}$)

here L denotes probability that outcome

1,2,3,4 (L = ${4 \over 6}$ = ${2 \over 3}$)

$\underline {} \,\,\,\underline {} \,\,\,\underline {} \,\,\underline 4 \,\,\underline 4 $

${1 \over {{6^2}}}\left( {{{{5^3}} \over {{6^3}}} + {{2{C_1}{{.5}^2}} \over {{6^3}}}} \right) = {{175} \over {{6^5}}}$

We can solve this problem by counting the number of "nice" subsets in the set S = {1, 2, $\ldots$, 20 }, and then dividing that number by the total number of possible subsets of S.

The sum of all elements in S is :

1 + 2 + $\ldots$ + 20 = $\frac{{20 \times 21}}{2}$ = 210

Since a "nice" subset must sum to 203, the elements not in the subset must sum to 210 - 203 = 7.

Now we need to find the ways to make the sum of 7 using the elements of S. The combinations are :

1. 1. 7
2. 2. 1 + 6
3. 3. 2 + 5
4. 4. 3 + 4
5. 5. 1 + 2 + 4
6. 6. 1 + 3 + 3(This doesn't work since 3 is repeated)
7. 7. 2 + 2 + 3(This doesn't work since 2 is repeated)

So, there are 5 "nice" subsets.

Since the set S has 20 elements, there are $2^{20}$ possible subsets (including the empty set and the set itself). The probability of randomly choosing a "nice" subset is therefore :

p (probability of getting white ball) = ${{30} \over {40}}$

q = ${1 \over 4}$ and n = 16

mean = np = 16.${3 \over 4}$ = 12

and standard diviation

= $\sqrt {npq} $ = $\sqrt {16.{3 \over 4}.{1 \over 4}} = \sqrt 3 $

$ \because $ $mean \over standard\,deviation$ = $12\over{\sqrt3}$ = $4{\sqrt3}$

Since sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space = ⁵C₂ + ⁶C₂

So, required probability = ${{{}^5{C_2}} \over {{}^5{C_2} + {}^6{C_2}}}$ = ${2 \over 5}$

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

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

${n_{\min }} = 5$

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

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)

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

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

It is given that there are three bags B₁, B₂ and B₃ and probabilities of being chosen B₁, B₂ and B₃ are respectively.

$ \therefore $ $P({B_1}) = {3 \over {10}},\,P({B_2}) = {3 \over {10}}$ and $P({B_3}) = {4 \over {10}}$.



Now, probability that the chosen ball is green, given that selected bag is

${B_3} = P\left( {{G \over {{B_3}}}} \right) = {3 \over 8}$

Now, probability that the selected bag is B₃, given that the chosen ball is green = $P\left( {{{{B_3}} \over G}} \right)$

$ = {{P\left( {{G \over {{B_3}}}} \right)P({B_3})} \over {P\left( {{G \over {{B_1}}}} \right)P({B_1}) + P\left( {{G \over {{B_2}}}} \right)P({B_2}) + P\left( {{G \over {{B_3}}}} \right)P({B_3})}}$

[by Baye's theorem]

= ${{\left( {{3 \over 8} \times {4 \over {10}}} \right)} \over {\left( {{5 \over {10}} \times {3 \over {10}}} \right) + \left( {{5 \over 8} \times {3 \over {10}}} \right) + \left( {{3 \over 8} \times {4 \over {10}}} \right)}}$

$ = {{{1 \over 2}} \over {{1 \over 2} + {5 \over 8} + {1 \over 2}}} = {4 \over {13}}$

Now, probability that the chosen ball is green = P(G)

= $P({B_1})P\left( {{G \over {{B_1}}}} \right) + P({B_2})P\left( {{G \over {{B_2}}}} \right) + P({B_3})P\left( {{G \over {{B_3}}}} \right)$

[By using theorem of total probability]

= $\left( {{3 \over {10}} \times {5 \over {10}}} \right) + \left( {{3 \over {10}} \times {5 \over 8}} \right) + \left( {{4 \over {10}} \times {3 \over 8}} \right)$

= ${3 \over {20}} + {3 \over {16}} + {3 \over {20}} = {{12 + 15 + 12} \over {80}} = {{39} \over {80}}$

Now, probability that the selected bag is B₃ and the chosen ball is green

= $P({B_3}) \times P\left( {{G \over {{B_3}}}} \right) = {4 \over {10}} \times {3 \over 8} = {3 \over {20}}$

Hence, options (a) and (c) are correct.

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 = ${{{2 \over 9}} \over {{2 \over 7} + {2 \over 9}}}$ = ${{2 \times 7} \over {18 + 14}}$ = ${7 \over {16}}$

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.